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EXERCISES , Questions 21 to 30
21. A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is µ. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip ?
ANSWER :: (a) Let Max Angular Speed be ω. At this angular speed the block will experience an outward force F= mω²L. This will be just equal to the frictional force = µmg. Equating the two we get,
mω²L=µmg
→ ω²=µg/L
→ ω=√(µg/L)
(b) When the speed is increased with an angular acceleration, the circular motion becomes non-uniform. So the block will have both radial and tangential accelerations.
The radial acceleration will be same as above =ω²L.
The tangential acceleration =dv/dt =d(ωL)/dt = L*dω/dt =Lα
So resultant acceleration will be
=√{(ω²L)²+(Lα)²}
= L√(ω4+α²)
So the force on it will be , mL√(ω4+α²) = µmg
→ ω4+α² =µ²g²/L² (Squaring both sides)
→ω =(µ²g²/L²-α²)1/4
22. A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure (7-E1). Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track. (a) Find the normal contact force by the road on the cycle when it is at B and at D. (b) Find the force of friction exerted by the track on the tyres when the cycle is at B, C and D. (c) Find the normal force between the road and the cycle just before and just after the cycle crosses C. (d) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed. ? Take g=10 m/s².
Figure for problem 22
ANSWER:: (a) B and D are crest and trough of the curvilinear road where the normals to the surface are vertical. So normal contact forces at these points by the road on the cycle will be vertically upwards balancing the effective weight of the cycle along with the rider.
Since the cycle has a uniform circular motion in the vertical plane, it will experience a radially outward force = mv²/r. At B its direction will be vertically upward opposite to weight. Thus effective weight of the cycle along with the rider at B
= mg-mv²/r
(m=100 kg, g=10m/s², r=100 m & v=18 km/h =18000/3600 m/s =5m/s)
= 100x10 -100x5²/100 N
= 1000-25 N = 975 N = Normal Contact force by the road on the cycle at B.
Effective weight of the cycle and rider at D
= mg+mv²/r
=100x10+100x5²/100 N
=1000+25 N
=1025 N = Normal contact force by the road on the cycle at D.
(b) At B and D the weight is balanced by the Normal Contact force and no force is on the cycle along the road. So there is no force of friction to oppose it at B and D. i.e. F = 0 N
At C the weight is acting at an angle of 45° to the surface. Resolving it along the surface and perpendicular to it we get a force equal to mg.cos45° acting on the cycle to accelerate it along the track. But the cycle does not accelerate, only goes along the track with a constant speed. So the force of friction is also equal to mg.cos45° (opposite to the direction of motion)
=100x10x1/√2 =500√2 N =707 N
(C) Normal force near the point C will be equal to the component of weight perpendicular to the surface = mg.sin45°
=100x10x1/√2 =500√2 N =707 N
Magnitude of Radial force just before and after the point C
=mv²/r = 100x5²/100 = 25 N
But before C due to convex nature of the surface, its direction will be opposite to weight component and reducing it to 707-25 N
=682 N = Normal force.
And after C due to the concave nature of the surface its direction will be along the weight component, thus total push on the road will be equal to 707+25 N = 732 N. Therefore Normal force here will also be = 732 N
(d) Let the minimum coefficient of friction be µ.
At C, Magnitudes of Frictional force is equal to =707 N,
Normal forces just before and after C are 682 N and 732 N respectively. So coefficient of friction µ at these points are 707/682 = 1.037 N and 707/732 =0.96
Though the lower value of µ is 0.96 but if we take this value just before C this will make available frictional force less than 707 N while the tangential component of weight remains 707 N. Thus it will skid and its speed won't be constant. So minimum coefficient of friction should be µ =1.037
23. In a children's park, a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod (Figure7-E2). Let the mass of each kid be 15 kg the distance between the points of the rod where the two kids hold it be 3.0 m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids.
Figure for problem 23
ANSWER :: Each kid needs and inward force to remain in uniform circular motion with the horizontal rod and this inward force is provided by the force of friction on the kid by the rod.
Value of this frictional force 'F' is, mω²r.
Here m=15 kg, r =3/2 m =1.5 m,
ω=20x2π/60 rads/s =2π/3 rads/s
Now F=15x(4π²/9)x1.5 = 10π² N.
24. A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle θ with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is µ. Find the range of the angular speed for which the block will not slip.
ANSWER :: Let the Angular Speed be ω, and the mass of the block be 'm'. See picture below,
Diagram for Poblem-24
For the circular motion of block, radius will be = R.sinθ
Outward horizontal force on the block F= mω²R.sinθ
Its tangential component = F.cosθ =mω²R.sinθ.cosθ
Normal component =F.sinθ =mω²R.sin²θ
Weight of the block =mg
Tangential component of weight = mg.sinθ
Normal component of weight = mg.cosθ
CASE-I
For minimum angular speed,
Net tangential force = mg.sinθ - mω²R.sinθ.cosθ
Net Normal force = mω²R.sin²θ+mg.cosθ
For the condition that block should not slip down,
mg.sinθ - mω²R.sinθ.cosθ = µ(mω²R.sin²θ+mg.cosθ)
→ mg.sinθ-µmg.cosθ = mω²R.sinθ.(cosθ+µ.sinθ)
→ g.sinθ-µg.cosθ = ω²R.sinθ.(cosθ+µ.sinθ)
→ ω² = g(sinθ-µ.cosθ)/R.sinθ.(cosθ+µ.sinθ)
→ ω = √[g(sinθ-µ.cosθ)/R.sinθ.(cosθ+µ.sinθ)]
CASE-II
For maximum angular speed
Net tangential force = mω²R.sinθ.cosθ -mg.sinθ
Net Normal force = mω²R.sin²θ+mg.cosθ
For the condition that block should not slip up,
mω²R.sinθ.cosθ -mg.sinθ =µ(mω²R.sin²θ+mg.cosθ)
→ mω²R.sinθ.(cosθ-µ.sinθ) =mg(sinθ+µcosθ)
→ ω²R.sinθ.(cosθ-µ.sinθ) =g(sinθ+µcosθ)
→ ω =√[g(sinθ+µcosθ)/R.sinθ.(cosθ-µ.sinθ)]
25. A particle is projected with a speed u at an angle θ with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circle? This radius is called radius of curvature of the curve at the point.
ANSWER:: You may remember such problems in projectile motion where the horizontal speed remains constant throughout its motion = u.cosθ. In this problem also near the highest position, same will be the speed, i.e. v = u.cosθ
If we consider a small part of its path as to be a circular arc, then the particle is moving with a uniform circular motion having this speed.
Take its mass as 'm', then inward force on this particle in circular motion is its weight, 'mg'. Now it can be related with its speed as,
mg = m(u.cosθ)²/R (Where R is the radius of curvature)
→ R = (u.cosθ)²/g
→ R = u².cos²θ/g
26. What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle θ/2 with the horizontal?
ANSWER :: Let us draw a diagram as below,
Diagram for Problem 26
Take the speed of the projectile at the given point be 'v'.
Since the horizontal velocity remains constant, so
v.cosθ/2 = u.cosθ
→ v = u.cosθ/(cosθ/2)
If we take its motion at the instant as uniform circular motion, we also need to know the centripetal force at the instant. It is provided by the component of the weight of the particle along the perpendicular to the speed v.
P = mg.cosθ/2
Now we can relate v and P as,
P = mv²/R
→ mg.cosθ/2 = mu².cos²θ/R(cos²θ/2)
→ R = u².cos²θ/g cos³(θ/2)
27. A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is µ. The block is given an initial speed v0 . As a function of the speed v write (a) the normal force by the wall on the block, (b)) the frictional force by the wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration {dv/dt=vdv/ds} to obtain the speed of the block after one revolution.
ANSWER:: Since the circular motion in this problem is not a uniform one, the block will have both the tangential and radial accelerations.
(a) Radial acceleration =v²/R (inward)
It is provided by the Normal force 'N' by the wall on the block.
So normal force by the wall on the block N = mv²/R
(b) The frictional force by the wall on the block,
F = µN
→ F = µmv²/R
(c) The tangential acceleration of the block is given by a=F/m.
Force of retardation on the block is frictional force 'F'
Since it is retardation, so we write
a = -F/m = -µmv²/mR = -µv²/R
(d) Since a = v.dv/ds
Speed after one revolution (Covering a distance s=2πR) is given by,
∫dv=∫(a/v).ds = ∫(-µv²/vR)ds = -µv/R∫ds
→ ∫(1/v)dv= (-µ/R)∫ds
{We integtrate dv from v0 to v and ds from 0 to 2πR for one revolution}
→ [ln v-ln v0] = (-µ/R)[2πR-0] = -2πµ
→ ln (v/v0) = -2πµ
→ v/v0 = e-2πµ
→ v = v0 e-2πµ
28. A table with a smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity ω in a circular path of radius R (figure 7-E3). A smooth groove AB of length L (<<R) is made on the surface of the table. The groove makes an angle θ with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move along AB. Find the time taken by the particle to reach the point B.
Figure for problem 28
ANSWER:: Since L<<R, we assume that force on the particle in the groove is constant from A to B.
Radial acceleration of the particle = ω²R
Component of acceleration along groove = ω²R.cosθ
Let it take time t to cover AB, using s=ut+½at²
L = 0+½ ω²R.cosθ.t² (Initial velocity u=0)
→ t² =2L/ω²R.cosθ
→ t =√{2L/ω²R.cosθ}
29. A car moving at a speed of 36 km /hr is taking a turn on a circular road of radius 50 m. A small wooden plate is kept on the seat with its plane perpendicular to the radius of the circular road (figure 7-E4). A small block of mass 100 g is kept on the seat which rests against the plate. The friction coefficient between the block and the plate is µ = 0.58. (a) Find the normal contact force exerted by the plate on the block. (b) The plate is slowly turned so that the angle between the normal to the plate and the radius of the road slowly increases. Find the angle at which the block will just start sliding on the plate.
Figure for problem 29
ANSWER:: Mass of the block, m = 100 g = 0.10 kg
v= 36 kmph =36000/3600 m/s =10 m/s
R = 50 m
(a) Normal force on the block by the plate = mv²/R
= 0.10 x 10²/50
= 0.20 N
(b) As the plate is turned, say by an angle θ, the normal force on the block by the plate decreases to a component of original normal force = mv²/R*cosθ N
Force of friction that will resist its sliding on the block will be
= µ*mv²/R*cosθ
While the force trying to slide it is = mv²/R*sinθ
At the point when the block will just start sliding, these two forces will be equal.
mv²/R*sinθ=µ*mv²/R*cosθ
tanθ = µ =0.58
→ θ = 30°
30. A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R (figure 7-E5). A smooth pully of small radius is fastened to the table. Two masses m and 2m placed on the table are connected through a string over the pulley. Initially, the masses are held by a person with the strings along the outward radius and then the system is released from the rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string.
ANSWER:: The system experiences an outward acceleration
= ω²R
Let tension in the string be T, m1 = m , m2 = 2m
If 'a' be the initial acceleration, For the first block,
T-mω²R = ma,
And for the second block,
2mω²R - T = 2ma,
Eliminating T in these two equations, we get
2mω²R - ma -mω²R =2ma
→ 3ma = mω²R
→ 3a = ω²R
→ a = ω²R/3
Put the value of 'a' in first equation,
T = mω²R+ mω²R/3
→ T= 4mω²R/3
===<<<O>>>===
Links for the chapter -
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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- 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
HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY
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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion
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HC Verma's Concepts of Physics, Chapter-4, The Forces
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