KKMishra's Tutorials: January 2018

Sunday, January 28, 2018

Solutions to Problems on "CENTER OF MASS, LINEAR MOMENTUM, COLLISION" - H C Verma's Concepts of Physics, Part-I, Chapter-9, QUESTIONS FOR SHORT ANSWER

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CENTER OF MASS, LINEAR MOMENTUM, COLLISION:--
QUESTIONS FOR SHORT ANSWER

1. Can the center of mass of a body be at a point outside the body?

Answer: Yes, the center of mass of a body can be outside the body. For example, the center of mass of a semi-circular wire does not lie inside the wire but at a point 2R/π above the center.

2. If all the particles of a system lie in X-Y plane, is it necessary that the center of mass be in X-Y plane?

Answer: Yes. Since Z-coordinate of each particle will be zero, hence ഽz.dm=0. So Z-coordinate of the center of mass Z=(1/M)ഽz.dm=0. So the center of mass lies in the X-Y plane.

3. If all the particles of a system lie in a cube, is it necessary that the center of mass be in the cube?

Answer: Yes. If we start from any two particles of this system, the center of mass (say A) of these two particles will lie on the line joining them. This line will be fully inside the cube. Next, these two particles can be replaced with their total mass at this center of mass A. Now the center of mass of third particle and previous two particles will lie on the line joining the third particle and A, Say this center of mass as B. Again this line joining A and third particle is fully inside the cube, so B is also inside the cube. The total mass of three particles can be assumed to be at B. In this way we can show that the center of mass of all the particles of a system that lies in a cube is necessarily in the cube.

4. The center of mass is defined as R=(1/M)Σimiri. Suppose we define "center of charge" as Rc =(1/Q)Σiqiri where qi represents the ith charge placed at ri and Q is the total charge of the system.

(a) can the center of charge of a two-charge system be outside the line segment joining the charges?

(b) If all the charges of a system are in X-Y plane, is it necessary that the center of charge be in X-Y plane?

Answer: (a) Yes. Because the charges are of two different type, positive and negative, unlike mass which is all similar type. For example, suppose that there is 2q charge at the origin and -q at 2 units along the x-axis. Its center of charge will be at a distance

= {1/(-q+2q)}*(-q*2+2q*0) =-2q/q =-2. Clearly, the center of charge, in this case, lies outside the line segment joining the two charges.

(b) Since the z-coordinate of all the charges, in this case, is zero so Σiqizi =0. Hence z-coordinate of center of charge =(1/Q)Σiqizi =0. It means the center of charge necessarily lies in X-Y plane.

(c) No. As we have seen in (a) that center of charge of two charges may lie outside the line segment joining the two charges, hence the center of charge, in this case, may lie outside the cube.

5. The weight Mg of an extended body is generally shown in a diagram to act through the center of mass. Does it mean that the earth does not attract other particles?

Answer: The earth does attract each and every particle. The resultant of all these forces of attraction is Mg which acts at the center of mass (where M is the total mass). That is why weight Mg of an extended body is generally shown in a diagram to act through the center of mass.

6. A bob suspended from the ceiling of a car which is accelerating on a horizontal road. The bob stays at rest with respect to the car with the string making an angle θ with the vertical. The linear momentum of the bob as seen from the road is increasing with time. Is it a violation of conservation of linear momentum? If not where is the external force which changes the linear momentum?

Answer: No, it is not a violation of conservation of linear momentum. The external force is the force of friction of road on the car which is acting through the string of the suspended bob. In the string, it is the tension force, component of which in horizontal direction changes the linear momentum.

7. You are waiting for a train on a railway platform. Your three-year-old niece is standing on your iron trunk containing the luggage. Why does the trunk not recoil as she jumps off on the platform?

Answer: The force applied by the girl on the trunk is less than the maximum static frictional force on the trunk by the platform.

8. In a head-on collision between two particles, is it necessary that the particles will acquire a common velocity at least for one instant?

Answer: Yes. Because during the collision the surfaces in contact will move equal distances in equal time interval as long as they remain in contact.

9. A collision experiment is done on a horizontal table kept in an elevator. Do you expect a change in the results if the elevator is accelerated up or down because of the noninertial character of the frame?

Answer: No. When we analyze the motion of a body with respect to a noninertial frame we add a pseudo force on the body opposite to the direction of acceleration of the frame. Since the elevator is accelerated only in vertical direction, the direction of the pseudo force is also vertical. It does not affect the horizontal movement of the body. So collision experiment done on a horizontal table in the elevator will not be affected if the elevator is accelerated up or down.

10. Two bodies make an elastic head-on collision on a smooth horizontal table kept in a car. Do you expect a change in the result if the car is accelerated on a horizontal road because of the noninertial character of the frame? Does the equation "Velocity of separation = Velocity of approach" remain valid in an accelerating car? Does the equation "final momentum = initial momentum" remain valid in the accelerating car?

Answer: In a noninertial frame both the bodies will experience equal acceleration in the horizontal direction. So the experiment result will change in the accelerating car.

After applying pseudo forces the motion can be analyzed as if in an inertial frame. Since the sum of external forces are not zero now, hence the equations (i) "Velocity of separation = Velocity of approach" and (ii) "final momentum = initial momentum" do not remain valid in the accelerating car.

11. If the total mechanical energy of a particle is zero, is its linear momentum necessarily zero? Is it necessarily nonzero?

Answer: (i) No.
Consider the mechanical energy of a particle under gravity. The total mechanical energy = K.E. + P.E.
K.E. is always positive because either v is positive or negative, v² will always be positive and so will ½mv² = K.E. but P.E. = mgh may be positive or negative because h is positive or negative depending upon its position with respect to the reference line. Suppose we take ground as the reference line. If the particle falls into a dry well, at some depth below ground level the magnitude of negative P.E. may be equal to K.E. thus making the total mechanical energy of the particle zero. But due to its velocity, the linear momentum is not zero.
(ii) No. It is not necessarily nonzero. Suppose the particle is on the ground at rest. K.E. = 0, and P.E. = 0. Total mechanical energy = 0. Since velocity is zero, the linear momentum is also zero.

12. If the linear momentum of a particle is known, can you find its kinetic energy? If the kinetic energy of a particle is known can you find its linear momentum?

Answer: (i) If only we know the mass of the particle, it's kinetic energy can be found out. Because both the K.E. and the magnitude of the momentum involve mass m and magnitude of the velocity v.
(ii) Even if we know the mass of the particle, only the magnitude of the momentum can be calculated. The direction of the momentum is unknown. Momentum, being a vector thus cannot be found out.

13. What can be said about the center of mass of a uniform hemisphere without making any calculation? Will its distance from the center be more than r/2 or less than r/2?

Answer: Less than r/2.
Just visualize a uniform cylinder of radius r and height r. Its center of mass will be at r/2 from the center of the base circle. Since the given hemisphere can be made by carving out masses from the upper parts the center of mass of the remaining hemisphere will be less than r/2.

14. You are holding a cage containing a bird. Do you have to make less effort if the bird flies from its position in the cage and manages to stay in the middle without touching the walls of the cage? Does it make a difference whether the cage is completely closed or it has rods to let air pass?

Answer: When the bird flies and maintains its position in the middle of the cage, it pushes air downward to get upward lift. The pushed air partly escapes down the rods of the cage, so the full downward pressure is not on the cage. So less effort will be needed to hold the cage.
If the cage is completely closed the full downward pressure of the pushed down air (equal to the weight of the bird) will be on the base of the cage. So same effort will be needed to hold the cage.

15. A fat person is standing on a light plank floating on a calm lake. The person walks from one end to the other on the plank. His friend sitting on the shore watches him and finds that the person hardly moves any distance because the plank moves backward about the same distance as the person moves on the plank. Explain.

Answer: Since the plank is light its mass is negligible in comparison to the fat person. Combined center of mass of the system containing the fat person and the plank can be considered near the person. Since there is no external force on this system the center of mass of the system will remain at the same place. Due to this the friend on the shore finds the fat person hardly move.

16. A high-jumper successfully clears the bar. Is it possible that his center of mass crossed the bar from below it? Try it with appropriate figures.

Answer: Nowadays high jumpers clear the rod keeping their center of mass at minimum possible height in order to put less effort. To do this they bend their body in a semicircular shape above the bar as shown in the figure below.

As we know the center of the mass of a uniform semicircular wire is at r-2r/π = (π-2)r/π = 0.36 r below its crest. So the center of mass may cross below the bar during the successful clearance of the bar as shown in the figure.

17. Which of the two persons shown in the figure (9-Q1) is more likely to fall down? Which external force is responsible for his falling down?

Figure for Q 17

Answer: Moving cart with an acceleration is a non-inertial frame for the passengers. So both of them will experience a pseudo force in the opposite direction of the movement. The person sitting on the left will be saved by the wall of the cart but not the person on the right. So the person on the right is more likely to fall and the force responsible for it is the pseudo force.

18. Suppose we define a quantity 'Linear Pomentum' as

Linear Pomentum = mass x speed.

The linear pomentum of a system of particles is the sum of linear pomenta of the individual particles. Can we state a principle of conservation of linear pomentum as "linear pomentum of a system remains constant if no external force acts on it"?

Answer: Let us consider a system of two particles of equal mass with equal speed but in opposite directions on a straight line. The center of mass of the system will remain stationary. So the speed of the center of mass = 0. Hence linear pomentum of the CoM = MxV =0. But from the definition linear pomentum of the system = mv+mv = 2mv. Hence we can not state such principle.
(Since speed v is scaler, we consider only the magnitude).

19. Use the definition of linear pomentum from the previous question. Can we state the principle of conservation of linear pomentum for a single particle?

Answer: If no external force is applied on the particle, the speed will remain the same. Hence we can state the conservation of linear pomentum for a single particle.

20. To accelerate a car we ignite petrol in the engine of the car. Since only an external force can accelerate the center of mass, is it proper to say that "the force generated by the engine accelerates the car"?

Answer: Technically it is not proper to say that the force generated by the engine accelerates the car because the engine does not generate a force along the acceleration. Even if it did so internal reaction force would have balanced it.
Since the movement of the piston of the engine makes a rotational motion in the wheels which in turn is used to push back the road and due to frictional force (External force) by the road, the car accelerates, hence in common language, such statement is made.

21. A ball is moved on a horizontal table with some velocity. The ball stops after moving some distance. Which external force is responsible for the change in the momentum of the ball?

Answer: Here frictional force is the external force which is responsible for the change in the momentum of the ball.

22. Consider the situation in the previous problem. Take "the table plus the ball" as the system. Friction between the table and the ball is then internal force. As the ball slows down the momentum of the system decreases. Which external force is responsible for the change in the momentum?

Answer: Here the external force responsible for the change in the momentum of the system is the frictional force by the ground on the table.

23. When a nucleus at rest emits a beta particle, it is found that the velocities of the recoiling nucleus and the beta particle are not along the same straight line. How can this be possible in in view of the principle of conservation of momentum?

Answer: The motion of subatomic particles are better explained with quantum physics. Even if we try to explain the given situation with Newtonian physics, it suggests that in view of the principle of conservation of momentum there must be additional subatomic particle released during the beta emission. As we know there is antineutrino also released during the beta emission.

24. A van is standing on a frictionless portion of a horizontal road. To start the engine the vehicle must be set in motion in the forward direction. How can the persons sitting inside the van do it without coming out and pushing from behind?

Answer: The persons sitting inside the van can move the van forward to start the engine without coming out if they simultaneously move backward. In absence of an external force like friction force by the road, the center of mass of van and persons together will remain stationary. So if the persons move backward simultaneously the van will move forward to keep the center of mass at the same position and the van will start.

25. In one dimensional elastic collision of equal masses, the velocities are interchanged. Can velocities in a one-dimensional collision be interchanged if the masses are not equal?

Answer: No, because the linear momentum will not be conserved.
For example, consider the one-dimensional collision of mass m with velocity 2v and mass 2m with velocity v before the collision. Total momentum = m.2v+2m.v = 4mv. Had the velocities been interchanged after the collision, final momentum of the system would be = m.v+2m.2v = 5mv, which is not according to the principle of conservation of the linear momentum. So the velocities, in this case, will not be interchanged.

===<<<O>>>===

Links to the chapter -
ALL LINKS

CHAPTER- 10 - Rotational Mechanics

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-15

EXERCISES - Q-16 TO Q-30

CHAPTER- 9 - Center of Mass, Linear Momentum, Collision

Questions for Short Answers

Objective - I

Objective - II

EXERCISES Q-1 TO Q-10

EXERCISES Q-11TO Q-20

EXERCISES Q-21 TO Q-30

EXERCISES Q-31 TO Q-42

EXERCISES Q-43 TO Q-54

HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY

Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)

HC Verma's Concepts of Physics, Chapter-7, Circular Motion

Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)

HC Verma's Concepts of Physics, Chapter-6, Friction

Click here for → Friction OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .

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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion

Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→ Newton's laws of motion - Objective - I

Click here for → Newton's Laws of Motion - Objective -II

Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

Click here for→Newton's Laws of Motion,Exercises(Q.No. 28 to 42)

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HC Verma's Concepts of Physics, Chapter-4, The Forces

The Forces-

"Questions for short Answers"

Click here for "The Forces" - OBJECTIVE-I

Click here for "The Forces" - OBJECTIVE-II

Click here for "The Forces" - Exercises

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HC Verma's Concepts of Physics, Chapter-3, Kinematics-Rest and Motion:---

Click here for "Questions for short Answers"

Click here for "OBJECTIVE-I"

Click here for EXERCISES (Question number 1 to 10)

Click here for EXERCISES (Question number 11 to 20)

Click here for EXERCISES (Question number 21 to 30)

Click here for EXERCISES (Question number 31 to 40)

Click here for EXERCISES (Question number 41 to 52)

===============================================

Chapter -2, "Vector related Problems"

Click here for "Questions for Short Answers"

Click here for "OBJECTIVE-I

Click here for "OBJECTIVE-II"

Click here for "Exercises"

Tuesday, January 16, 2018

Solutions to Problems on "Work and Energy" - H C Verma's Concepts of Physics, Part-I, Chapter-8, EXERCISES Q-55 TO Q-64

My Channel on YouTube → SimplePhysics with KK

For links to

other chapters - See bottom of the page

Or click here → kktutor.blogspot.com

WORK AND ENERGY:--
EXERCISES (55-64)

55. Figure (8-E15) shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring of spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal.

Figure for Q 55

Answer: Let the required compression of the spring be x.

P.E. stored in the spring = ½kx²

When the spring is released, the spring does a work on the block and changes its K.E. which is equal to the P.E. stored in the spring = ½kx². When the block reaches P, a part of this K.E. is converted to P.E. Let v be the velocity of the block at P.

It's energy at P =K.E.+P.E.

= ½mv²+mgR (Its height above flat surface = R)

Hence, ½mv²+mgR = ½kx²

→mv²+2mgR =kx² ------------------- (i)

Since the block presses the track with a force mg, that means Normal force on the block is also mg which due to its velocity should be mv²/R. Hence, mv²/R = mg

→mv² =mgR

Putting this value in (i), we get,

mgR+2mgR = kx²

→kx² = 3mgR

→x² = 3mgR/k

→x = √(3mgR)/k

56. The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of √(3gl). Find the angle rotated by the string before it comes slack.

Answer: Assuming l = length of the string. Speed given to the bob = u = √(3gl)

K.E. given to the bob = ½mu² = ½m*3gl =3mgl/2 ---- (i)

The string will become slack at the point where the tension in the string T = 0. Suppose this point comes after rotating an angle θ and bob's speed here is v. If at this position the string makes an angle ⲫ with the upward vertical then ⲫ = 180°-θ. Its height from the original position A is = h. (See the figure below.)

Figure for Q 56

h = AC + DC = l + l.cosⲫ = l(1 + cosⲫ)

Taking the level of original position of the bob as zero P.E. level, P.E. of the bob at final position = mg(l+l.cosⲫ)

=mgl(1+cosⲫ)

And K.E. = ½mv²

Total mechanical energy at this point = ½mv²+mgl(1+cosⲫ)

Since total mechanical energy will be conserved, hence (i) and (ii) will be equal.

Equating (i) and (ii) we get,

½mv²+mgl(1+cosⲫ) = 3mgl/2

→v² + 2gl(1+cosⲫ) = 3gl ------------- (iii)

Consider the forces at the final position. Force along the string towards center = T+mg.cosⲫ and centripetal acceleration mv²/l. So,

T+mg.cosⲫ = mv²/l

→mv²/l = mg.cosⲫ (since here T=0)

→v² = gl.cosⲫ

Putting this value in (iii)

gl.cosⲫ + 2gl(1+cosⲫ) = 3gl

→gl.cosⲫ + 2gl.cosⲫ = 3gl - 2gl = gl

→3gl.cosⲫ = gl

→3.cosⲫ = 1

→cosⲫ = 1/3

→cos(180°-θ) = 1/3

→ -cosθ = 1/3

→ cosθ = - 1/3

→θ = cos⁻¹(-1/3)

57. A heavy particle is suspended by a 1.5 m long string. It is given a horizontal velocity of √57 m/s. (a) Find the angle made by the string with the upward vertical, when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take g =10 m/s².

Answer: Let us draw the figure for this problem.

Figure for Q 57

(a) u=√57 m/s, K.E. = ½mu² = 57m/2 J

As in the problem 56, at the point of string being slack,

mv²/l = mg.cosⲫ

→v² = lg.cosⲫ

Total energy at final point = K.E. + P.E.

= ½mv² + mgl(1+cosⲫ)

=½mlg.cosⲫ + mgl(1+cosⲫ) = 57m/2

(Since final energy will be equal to the initial energy)

→ lg.cosⲫ + 2gl(1+cosⲫ) = 57

→ 3gl.cosⲫ = 57 - 2gl

→ 3*10*1.5.cosⲫ = 57 - 2*10*1.5 =27

→ 45.cosⲫ = 27

→ cosⲫ = 27/45 = 3/5 = 0.60 = cos53°

→ ⲫ = 53°

(b) We have v² = lg.cosⲫ = 1.5*10*3/5 = 9

→ v = 3.0 m/s

(c) Once the string becomes slack, the bob will move like a projectile with initial velocity = 3.0 m/s and angle of projection ⲫ= 53°. For this projectile motion maximum height above the point of projection d = (v.sinⲫ)²/2g.

Let the maximum height reached above the point of suspension = x.

x = l.cosⲫ + d

= l.cosⲫ + (v.sinⲫ)²/2g

=1.5*3/5 + (3*4/5)²/2*10

= 0.9 + 12*12/5*5*20 =0.9 + 0.288 = 1.188

→x ≈ 1.2 m/s

58. A simple pendulum of length L having a bob of mass m is deflected from its rest position by an angle θ and released (Figure 8-E16). The string hits a peg which is fixed at a distance x below the point of suspension and the bob starts going in a circle centered at the peg. (a) Assuming that initially, the bob has a height less than the peg, show that the maximum height reached by the bob equals its initial height. (b) If the pendulum is released with θ = 90° and x = L/2 find the maximum height reached by the bob above its lowest position before the string becomes slack. (c) Find the minimum value of x/L for which the bob goes in a complete circle about the peg when the pendulum is released from θ = 90°.

Figure for Q 58

Answer: (a) Let the initial height of the bob above the rest position be h. Taking the rest position as zero P.E. level, Its initial P.E. = mgh.

Initial K.E. = 0 because speed is zero. So total Energy = mgh.

Figure 58 (a)

When the bob reaches the maximum height say h', its P.E. = mgh' and K.E. = 0 because again speed is zero. Now total energy = mgh'.

Since total energy will be conserved.

mgh' = mgh

→h' = h

Hence proved.

(b) Taking the lowest position of the bob as zero potential energy level. At the point θ = 90°, P.E. of the bob = mgL, K.E. =0, Hence total energy = mgL.

Figure for 58b

Let at the point where string becomes slack, its speed = v and the string makes θ with the upward vertical. Toatal energy at this point = K.E. + P.E.

= ½mv² + mg(L/2+L/2*cosθ)

= ½mv²+mgL/2*(1+cosθ)

Total energy will remain the same, hence

½mv²+mgL/2*(1+cosθ) = mgL

½v²+gL/2*(1+cosθ) = gL

v²+gL*(1+cosθ) = 2gL --------------- (i)

Here tension in the string T = 0 and only centripetal force = mg.cosθ.

Hence mv²/½L = mg.cosθ

→v² = (gL/2)*cosθ

Putting it in (i)

(gL/2)*cosθ +gL*(1+cosθ) = 2gL

→ ½cosθ +(1+cosθ) = 2

→ 3/2*cosθ = 1

→ cosθ = 2/3

So maximum height reached by the bob above the lowest point

=L/2 + L/2*cosθ

=L/2*(1+2/3)

=L/2*(5/3)

=5L/6

(c) When the bob makes a complete circle in the above condition, the weight of the bob is the only centripetal force on the bob. Hence mv²/(L-x) = mg

→v² = g(L-x)

Figure for 58c

Equating the energies at top of the circle and at the initial point.

mg*2(L-x) + ½mv² = mgL

→ mg*2(L-x) + ½mg(L-x) = mgL

→2(L-x) + (L-x)/2 = L

→5(L-x)/2 = L

→5(L-x) = 2L

→5L-5x = 2L

→3L = 5x

→5x = 3L

→x/L = 3/5 = 0.60

59. A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere.

Answer: Let the radius to the particle make an angle θ with the upward vertical. If the normal force on the block is N and weight mg then centripetal force = mg.cosθ-N.

If the velocity of the particle be v, then

mg.cosθ-N = mv²/r (Where r is the radius of the sphere)

At the point where the particle leaves the surface, the normal force N = 0. So,

mg.cosθ = mv²/r

→v² = gr.cosθ

Figure for Q 59

Change in height of the particle = r-r.cosθ

Change in P.E. at the point it leaves the surface = mg(r-r.cosθ)

This change in P.E. will result in increase of K.E. =½mv²

So, mg(r-r.cosθ)=½mv² =½mgr.cosθ

→1-cosθ=½cosθ

→(3/2)*cosθ=1

→cosθ=2/3

→θ=cos⁻¹(2/3)

60. A particle of mass m is kept on a fixed smooth sphere of radius R at a position where the radius through the particle makes an angle 30° with the vertical. The particle is released from the position. (a) what is the force exerted by the sphere on the particle just after the release? (b) find the distance traveled by the particle before it leaves contact with the sphere.

Answer: (a) Since the sphere is smooth, the force exerted by the sphere on the particle will be normal to the surface =N. Weight of the particle =mg (Downward)

Equating the force along the radius, N=mg.cos30°
=mg.√3/2
=√3mg/2
(b) Let it travel through an angle θ before it leaves the surface. At this point N=0. If its velocity at this point is v, then
mg.cos(θ+30°) = mv²/R
→ g.cos(θ+30°) = v²/R
→v² = gR.cos(θ+30°) ------------------------ (i)
Change in P.E. = mg{Rcos30°-Rcos(θ+30°)}
Change in K.E. =½mv²
Equating these two
½mv² = mg{Rcos30°-Rcos(θ+30°)}
→v² = 2g{Rcos30°-Rcos(θ+30°)}
Putting the value of v² from (i)
gR.cos(θ+30°)=2g{Rcos30°-Rcos(θ+30°)}
→cos(θ+30°) = 2cos30° - 2cos(θ+30°)
→3cos(θ+30°) = √3
→cos(θ+30°) = √3/3 = 1/√3 ≈cos55°
→θ = 55°-30° = 25° = 0.43 radian
Hence distance traveled by the particle = Rθ = 0.43R

61. A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontal speed v. (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of v for which the particle does not slip on the sphere? (c) Assuming the velocity v to be half the minimum calculated in part (b), find the angle made by the radius through the particle with the vertical when it leaves the sphere.

Answer: (a) At the top weight of the particle = mg (downward)

Normal force on the particle = N (Upward)

Net centripetal force = mg - N

Centripetal acceleration = mv²/R

Hence mg-N = mv²/R

→N = mg-mv²/R

(b) For the limiting state when the particle is just to leave the surface N = 0

Hence 0 = mg - mv²/R

→v²/R = g

→v² = gR

→v=√(gR)

(c) Given that v = ½√gR

K.E. = ½mv² = ½m*gR/4 = mgR/8

At the point of leaving the surface, the particle descends through R-R.cosθ. (Where θ is the angle by the radius through particle and upward vertical)

The decrease in P.E. = mg(R-R.cosθ)

If the speed of the particle at this point is u, its K.E. =½mu²

Equating the energies at initial and final point,

½mu² = mgR/8 + mg(R-R.cosθ)
→u² = gR/4 + 2gR-2gR.cosθ = 9gR/4-2gR.cosθ ------- (i)
At this point N = 0.
Centripetal force = mg.cosθ
Hence, mu²/R = mg.cosθ
u² = gR.cosθ
Putting it (i)
gR.cosθ = 9gR/4 - 2gR.cosθ
→3cosθ = 9/4
→cosθ = 3/4
→θ = cos⁻¹(3/4)

62. Figure (8-E17) shows a smooth track which consists of a straight inclined part of length l joining smoothly with the circular part. A particle of mass m is projected up the incline from its bottom. (a) Find the minimum projection-speed v₀ for which the particle reaches the top of the track. (b) Assuming that the projection-speed is 2v₀ and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the projection-speed is only slightly greater than v₀, where will the block lose contact with the track?

Figure for Q 62

Answer: (a) Height gained in reaching the Top

H=l.sinθ+h =l.sinθ+R-R.cosθ

P.E. at the top = mgH = mg(l.sinθ+R-R.cosθ)

K.E. at top = 0

Total energy at the top = P.E.+K+E.

= mg(l.sinθ+R-R.cosθ)

Initial P.E. = 0

Initial K.E. = ½mv₀²
Initial total energy = ½mv₀²
Equating these two
½mv₀² = mg(l.sinθ+R-R.cosθ)

Figure for Q 62

v₀² = 2g(l.sinθ+R-R.cosθ) =2g{l.sinθ+R(1-cosθ)}

v₀ = √[2g{l.sinθ+R(1-cosθ)}]

(b) Initial speed = 2v₀

Let the final speed = v

Total energy at the top =Toatal energy at the initial point

½mv²+mgH = ½m(2v₀)² =2mv₀²

→v²+2gH = 4v₀²

→v² = 4v₀²-2gH

Centripetal acceleration = v²/R
Hence force on the particle = mv²/R =(m/R)*(4v₀²-2gH)
=(m/R)*[8g{l.sinθ+R(1-cosθ)}-2g*(l.sinθ+R-R.cosθ)]
=(m/R)*6g{l.sinθ+R(1-cosθ)}
=6mg(1-cosθ+l.sinθ/R)
(c) If the initial speed is slightly more than v₀, it means the block just starts from the rest at top. Let the radius through the point at which the block leaves the surface makes an angle θ with the vertical and its velocity there is u. Normal force here = 0. Hence,
mg.cosθ = mu²/R
→u² = gR.cosθ
K.E. =½mu², Change in P.E. =mg(R-R.cosθ). Equating,
u² = 2gR(1-cosθ)
or, gR.cosθ = 2gR(1-cosθ)
→cosθ = 2-2.cosθ
→3.cosθ = 2
→cosθ = 2/3
→θ = cos⁻¹(2/3)

63. A chain of length l and mass m lies on the surface of a smooth sphere of radius R>l with one end tied to the top of the sphere. (a) Find the gravitational potential energy of the chain with reference level at the center of the sphere. (b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle θ. (c) Find the tangential acceleration dv/dt of the chain when the chain starts sliding down.

Answer: (a) See the figure below.

Figure for Q 63(a)

Let the angle subtended by the chain at center = α
Length of the chain = l
Hence α = l/R
Let us take a very small length of the chain (at angle θ from the vertical) that subtends angle dθ at the center.
Length of this small length =Rdθ
Mass of this small length = (m/l)*Rdθ
Height of this small length above the center = R.cosθ
P.E. of this small length = (m/l)*Rdθ*g*R.cosθ
= (mR²g/l)*cosθdθ
Integrating this over an angle α we get P.E. of the chain.
=∫(mR²g/l)*cosθdθ = (mR²g/l)*∫cosθdθ = (mR²g/l)[sinθ]₀α
= (mR²g/l)sinα = (mR²g/l)sin(l/R)

(b) K.E. of the chain = Change in P.E. of the chain.
To know the P.E. of the chain after it slides down an angle θ, we shall put the limits of integration from θ to β. Where these are the angles made by the radii touching the ends of the chain. β = θ+α
So P.E. at this point = (mR²g/l)[sinθ]^βθ = (mR²g/l)[sinβ-sinθ]
= (mR²g/l)[sin(θ+α)-sinθ]
= (mR²g/l)[sin(θ+l/R)-sinθ]
Initial P.E. of the chain = (mR²g/l)sin(l/R)
Change in P.E. of the chain
= (mR²g/l)sin(l/R) - (mR²g/l)[sin(θ+l/R)-sinθ]
= (mR²g/l)[sin(l/R) - sin(θ+l/R)+sinθ] =K.E. of the chain after it slid through an angle θ.
(c) As we derived the expression for K.E. of the chain at an instant when slid angle = θ, we have K.E. =½mv² (where v is speed at this instant). So we have,
½mv² = (mR²g/l)[sin(l/R) - sin(θ+l/R)+sinθ] ----(i)
Differentiating it with respect to time t,
→ ½m(2vdv/dt) = (mR²g/l)[0-cos(θ+l/R).dθ/dt+cosθ.dθ/dt]
→ vdv/dt = (R²g/l) [-cos(θ+l/R).dθ/dt+cosθ.dθ/dt]
→ dv/dt = (R²g/l) [-cos(θ+l/R).dθ/dt+cosθ.dθ/dt]/v
→dv/dt=(R²g/l)[-cos(θ+l/R).dθ/dt+cosθ.dθ/dt]/Rdθ/dt
→dv/dt=(Rg/l)[cosθ-cos(θ+l/R)]
→dv/dt=(Rg/l)[cos0-cos(0+l/R)] (when chain starts sliding θ=0)
→dv/dt=(Rg/l)[1-cos(l/R) =Tangential acceleration

64. A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides.

Answer: Since the sphere is moving with a constant acceleration a, it is a non-inertial frame for the particle. When analyzing the particle we should apply a pseudo force ma on the particle opposite to the direction of a. When at angle θ, forces on it are mg downward and pseudo force ma horizontal and normal force N radial outward as shown in the figure.