KKMishra's Tutorials: November 2016

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EXERCISES , Questions 1 to 10

1. Find the acceleration of the moon with respect to the earth from the following data: Distance between the earth and the moon = 3.85x10⁵ km and the time taken by the moon to complete one revolution around the earth =27.3 days.

ANSWER :: T = Time period = 27.3 days

=27.3 x 24 x 60 x 60 s =2358720 s

Angular speed, ω = 2π /T =2π/2358720 rad/s

R = 3.85x10⁵ km = 3.85x10⁸ m

In a uniform circular motion, we have only radial acceleration, here
Acceleration of the moon with respect to earth

= (ω)²R

= (2π/2358720)²x 3.85x10⁸ m/s²

=4x3.14x3.14x 3.85x10⁸ / (2358720x2358720) m/s²

=0.002729 m/s²

= 2.73x10^-3 m/s².

2. Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of the earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.

ANSWER :: T= Time period = 24 hrs =24x3600 s

ω = 2π /T =2π/(24x3600) rad/s

Diameter of earth =12800 km

→ Radius of earth, R=12800/2 =6400 km =64x10⁵ m

Radial acceleration = (ω)²R

= 4π²x64x10⁵/(24x3600)² m/s²

=0.0338 m/s².

3. A particle moves in a circle of radius 1.0 cm at a speed given by v =2.0t where v is in cm/s and t in seconds.

(a) Find the radial acceleration of the particle at t =1 s. (b) Find the tangential acceleration at t=1 s (c) Find the magnitude of the acceleration at t = 1 s.

ANSWER :: (a) At t =1 s, v =2.0 cm/s.

So radial acceleration a_r = v²/r = 4/1 = 4.0 cm/s².

(b) Tangential acceleration a_t = dv/dt =d(2.0t)/dt =2.0 cm/s².

(c) Magnitude of Acceleration = √(a_r²+a_t² ) =√(4²+2²) =√20 cm/s².

4. A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What horizontal force on the scooter is needed to make the turn possible?

ANSWER :: Mass, m=150 kg ,

Speed, v=36 km/hr =36000 m/3600 s =10 m/s

Radius, r = 30 m.

Force needed to make the turn possible = mv²/r

=150x10x10/30 = 500 N.

5. If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?

ANSWER:: If the horizontal force needed for the turn is to be supplied fully by the normal force (i.e. no contribution from frictional force) then,

N.sinθ = mv²/r, (N= Normal force, θ = angle of banking) →N.sinθ =500 ............. (i)

And, N.cosθ =mg,

→N.cosθ=150x10=1500 ....... (ii), (taking g=10 m/s²)

See the picture below,

Diagram for the problem no 5

Dividing (i) by (ii) we get,

tanθ =500/1500 =1/3

So, θ =tan-1(1/3).

6. A park has a radius of 10 m. If a vehicle goes around it at an average speed of 18 km/hr, what should be the proper angle of banking?

ANSWER:: As in the previous problem

N.sinθ=mv²/r ----- (i)

N.cosθ=mg ------(ii)

Dividing (i) by (ii)

tanθ =v²/gr

In this problem r =10 m, v =18 km/hr =18000/3600=5 m/s,

(take g =10 m/s²)

So, tanθ =25/(10x10)

=25/100

=1/4

→ θ =tan-1(1/4).

7. If the road of the previous problem is horizontal (No Banking), what should be the minimum friction coefficient so that a scooter going at 18 km /hr does not skid?

ANSWER:: In such case, minimum frictional force = Centripetal force

→ µN =mv²/r

→ µ.mg =mv²/r (∵ N=mg)

→ µ =v²/gr

=5²/(10x10)

=25/100

=1/4

=0.25

8. A circular road of radius 50 m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used?

ANSWER :: As in the case of problem 6 where no frictional force is used to turn,

tanθ =v²/gr → v²=gr.tanθ

here, r =50 m, θ =30° & take g =10 m/s²

it gives v² =10x50xtan30° =500x1/√3 =500/1.732 =288.68≈289

→v =17 m/s.

So, at 17 m/s, a vehicle would not require friction to turn on this road.

9. In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the center at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coulomb attraction. In the ground state, the electron goes around the proton in a circle of radius 5.3x10-11 m. Find the speed of the electron in the ground state. Mass of the electron = 9.1x10-31 kg and charge of the electron = 1.6x10-19 C.

ANSWER :: Here r =5.3x10-11 m

m =9.1x10-31 kg

Charge on electron or proton, q =1.6x10-19 C

v =speed of electron =?

Force of Coulomb attraction on the electron =kq²/r²

(constant, k =9x109 m²/C² )

This coulomb attraction force is the centripetal force =mv²/r giving the electron required turn on the circular path, thus

mv²/r =kq²/r²

→v² =kq²/mr

= 9x109 x(1.6x10-19)²/(9.1x10-31 x5.3x10-11)

=0.478 x1013

=4.78 x1012

→v =2.186x106 m/s

≈2.2x106 m/s.

10. A stone is fastened to one end of a string and is whirled in a vertical circle of radius R. Find the minimum speed of the stone can have at the highest point of the circle.

ANSWER:: At the highest point of the circle, the forces acting on the stone are, its weight mg, and the tension in the string, T both in the downward direction. So the total force on the string is

T +mg in the downward direction towards the center of the circle. See the diagram below:--

Diagram for Q-10

This force acts as the centripetal force on the stone that maintains its circular path. Due to this, the stone has a radial acceleration,

a = v²/R.

So from Newton's Laws of Motion,

Force = mass * acceleration

→ T +mg =mv²/R

Where 'm' is the mass of the stone, 'v' is the speed of the stone and 'R' is the radius of the circular path. Now for the stone to have the minimum speed 'v', the value of the expression on the left-hand side should be the minimum. Since mg is fixed, the value of the left-hand side will be the minimum when T

In such a case,

mg = mv²/R

→v² =gR

→v =√(gR).

So at the highest point of the circle, the stone can have the minimum speed equal to √(gR).

===<<<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

CHAPTER-8 - Work and Energy

Questions for Short Answers

Objective - I

Objective - II

Exercises - Q 1 to Q 10

Exercises - Q 11 to Q 20

Exercises - Q 21 to Q 30

Exercises - Q 31 to Q 42

Exercises - Q 43 to Q 54

Exercises - Q 55 to Q 64

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)

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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Wednesday, November 16, 2016

My Channel on You Tube ---> SimplePhysics with KK

For links to

other chapters - See bottom of the page

EXERCISES , Questions 1 to 10

1. Find the acceleration of the moon with respect to the earth from the following data: Distance between the earth and the moon = 3.85x105 km and the time taken by the moon to complete one revolution around the earth =27.3 days.

ANSWER :: T = Time period = 27.3 days

=27.3 x 24 x 60 x 60 s =2358720 s

Angular speed, ω = 2π /T =2π/2358720 rad/s

2. Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of the earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.

ANSWER :: T= Time period = 24 hrs =24x3600 s

ω = 2π /T =2π/(24x3600) rad/s

Diameter of earth =12800 km

→ Radius of earth, R=12800/2 =6400 km =64x105 m

Radial acceleration = (ω)²R

= 4π²x64x105/(24x3600)² m/s²

=0.0338 m/s².

3. A particle moves in a circle of radius 1.0 cm at a speed given by v =2.0t where v is in cm/s and t in seconds.

(a) Find the radial acceleration of the particle at t =1 s. (b) Find the tangential acceleration at t=1 s (c) Find the magnitude of the acceleration at t = 1 s.

ANSWER :: (a) At t =1 s, v =2.0 cm/s.

So radial acceleration ar = v²/r = 4/1 = 4.0 cm/s².

(b) Tangential acceleration at = dv/dt =d(2.0t)/dt =2.0 cm/s².

(c) Magnitude of Acceleration = √(ar²+at² ) =√(4²+2²) =√20 cm/s².

4. A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What horizontal force on the scooter is needed to make the turn possible?

ANSWER :: Mass, m=150 kg ,

Speed, v=36 km/hr =36000 m/3600 s =10 m/s

Radius, r = 30 m.

Force needed to make the turn possible = mv²/r

=150x10x10/30 = 500 N.

5. If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?

ANSWER:: If the horizontal force needed for the turn is to be supplied fully by the normal force (i.e. no contribution from frictional force) then,

N.sinθ = mv²/r, (N= Normal force, θ = angle of banking) →N.sinθ =500 ............. (i)

And, N.cosθ =mg,

→N.cosθ=150x10=1500 ....... (ii), (taking g=10 m/s²)

See the picture below, Diagram for the problem no 5

Diagram for the problem no 5

Dividing (i) by (ii) we get,

tanθ =500/1500 =1/3

So, θ =tan-1(1/3).

6. A park has a radius of 10 m. If a vehicle goes around it at an average speed of 18 km/hr, what should be the proper angle of banking?

ANSWER:: As in the previous problem

N.sinθ=mv²/r ----- (i)

N.cosθ=mg ------(ii)

Dividing (i) by (ii)

tanθ =v²/gr

In this problem r =10 m, v =18 km/hr =18000/3600=5 m/s,

(take g =10 m/s²)

7. If the road of the previous problem is horizontal (No Banking), what should be the minimum friction coefficient so that a scooter going at 18 km /hr does not skid?

ANSWER:: In such case, minimum frictional force = Centripetal force

→ µN =mv²/r

→ µ.mg =mv²/r (∵ N=mg)

→ µ =v²/gr

=5²/(10x10)

=25/100

=1/4

=0.25

8. A circular road of radius 50 m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used?

ANSWER :: As in the case of problem 6 where no frictional force is used to turn,

tanθ =v²/gr → v²=gr.tanθ

ANSWER :: Here r =5.3x10-11 m

m =9.1x10-31 kg

Charge on electron or proton, q =1.6x10-19 C

v =speed of electron =?

Force of Coulomb attraction on the electron =kq²/r²

(constant, k =9x109 m²/C² )

This coulomb attraction force is the centripetal force =mv²/r giving the electron required turn on the circular path, thus

mv²/r =kq²/r²

→v² =kq²/mr

= 9x109 x(1.6x10-19)²/(9.1x10-31 x5.3x10-11)

=0.478 x1013

=4.78 x1012

→v =2.186x106 m/s

10. A stone is fastened to one end of a string and is whirled in a vertical circle of radius R. Find the minimum speed of the stone can have at the highest point of the circle.

ANSWER:: At the highest point of the circle, the forces acting on the stone are, its weight mg, and the tension in the string, T both in the downward direction. So the total force on the string is

T +mg in the downward direction towards the center of the circle. See the diagram below:--

Diagram for Q-10

This force acts as the centripetal force on the stone that maintains its circular path. Due to this, the stone has a radial acceleration,

a = v²/R.

So from Newton's Laws of Motion,

Force = mass * acceleration

1. Find the acceleration of the moon with respect to the earth from the following data: Distance between the earth and the moon = 3.85x10⁵ km and the time taken by the moon to complete one revolution around the earth =27.3 days.

→ Radius of earth, R=12800/2 =6400 km =64x10⁵ m

= 4π²x64x10⁵/(24x3600)² m/s²

So radial acceleration a_r = v²/r = 4/1 = 4.0 cm/s².

(b) Tangential acceleration a_t = dv/dt =d(2.0t)/dt =2.0 cm/s².

(c) Magnitude of Acceleration = √(a_r²+a_t² ) =√(4²+2²) =√20 cm/s².

See the picture below,

Diagram for the problem no 5