Rotation Rock And Roll(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

ROTATION ROCK AND ROLL MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. The sphere shown in figure lies on a rough plane when a particle of mass m travelling at a speed v0 collides and sticks with it. If the line of motion of the particle is at a distance h above the plane, find the linear speed of the combined system just after the collision. Assume that the mass M of the sphere is large compared to the mass of the particle so that the centre of mass of the combined system is not appreciably shifted from the centre of the sphere.

2mv0M+m
Mv0M+m
V0
mv0M+m

Discuss Question

Answer: Option D. -> mv0M+m
:
D
Take the particle plus the sphere as the system. Using conservation of linear momentum, the linear speed of the combined system v is given by

m v_{0} = (M + m) v o r, v = \frac{m v_{0}}{M + m}

. .....(i)

Question 12. Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia

1.6 \times 10^{- 4} k g - m^{2}

and a radius 2.0 cm. Find 'x', which is the kinetic energy of the system as the 400 g block falls through 50 cm and also find 'y', the speed of the blocks at this instant.

x = 10.5 J, y = 2.6 m/s
x = 9.8 J, y = 2.6 m/s
x = 9.8 J, y = 1.4 m/s
x = 0.98 J, y = 1.4 m/s

Discuss Question

Answer: Option D. -> x = 0.98 J, y = 1.4 m/s
:
D
According to the question

Two Blocks Of Masses 400 G And 200 G Are Connected Through A...

0.4 g - T_{1} = 0.4 a

...(1)

T_{2} - 0.2 g = 0.2 a

....(2)

(T_{1} - T_{2}) r = \frac{I a}{r}

...(3)
From equation 1,2 and 3

\Rightarrow a = \frac{(0.4 - 0.2) g}{(\frac{0.4 + 0.2 + 1.6}{0.4})} = \frac{g}{5}

Therefore (y) V=

\sqrt{2 a h} = \sqrt{(2 \times g l^{5} \times 0.5)}

\Rightarrow \sqrt{(\frac{g}{5})} = \sqrt{(\frac{9.8}{5})} = \frac{1.4 m}{s}

.
(x) total kinectic energy of the system

= \frac{1}{2} m_{1} V^{2} + \frac{1}{2} m_{2} V^{2} + \frac{1}{2} 18^{2}

= (\frac{1}{2} \times 0.4 \times {1.4}^{2}) + (\frac{1}{2} \times 0.2 \times {1.4}^{2}) + (\frac{1}{2} \times (\frac{1.6}{4}) \times {1.4}^{2}) = 0.98 j o u l e

Question 13. A block of mass 2 kg hangs from the rim of a wheel of radius 0.5 m. On releasing 2 kg from rest the block falls through 5 m height in 2 s. The moment of inertia of the wheel will be

1 kg−m2
3.2 kg−m2
2.5 kg−m2
1.5 kg−m2

Discuss Question

Answer: Option D. -> 1.5 kg−m2
:
D
On releasing from rest the block falls through 5 m height in 2 sec.

5 = 0 + \frac{1}{2} a (2)^{2} [A s S = u t + \frac{1}{2} a t^{2}] ∴ a = 2.5 m / s^{2}

Substituting the value of a in the formula

a = \frac{g}{1 + \frac{I}{m R^{2}}}

and by solving we get

\Rightarrow 2.5 = \frac{10}{1 + \frac{I}{2 \times (0.5)^{2}}} \Rightarrow I = 1.5 k g - m^{2}

Question 14. Two discs of moment of inertia

I_{1}

and

I_{2}

and angular speeds

ω_{1}

and

ω_{2}

are rotating along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate together along the same axis the rotational KE of system will be

I1ω1+I2ω22(I1+I2)
(I1ω1+I2ω2)22(I1+I2)
(I1+I2)(ω1+ω2)22
None of these

Discuss Question

Answer: Option B. -> (I1ω1+I2ω2)22(I1+I2)
:
B
By the law of conservation of angular momentum

I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω

Angular velocity of system

ω = \frac{I_{1} ω_{1} + I_{2} ω_{2}}{I_{1} + I_{2}}

Rotational kinetic energy

= \frac{1}{2} (I_{1} + I_{2}) ω^{2} = \frac{1}{2} (I_{1} + I_{2}) {(\frac{I_{1} ω_{1} + I_{2} ω_{2}}{I_{1} + I_{2}})}^{2} = \frac{(I_{1} ω_{1} + I_{2} ω_{2})^{2}}{2 (I_{1} + I_{2})} .

Question 15. A thin circular ring of mass 'M' and radius 'R' is rotating about its axis with a constant angular velocity

ω

. Four objects each of mass 'm', are kept gently to the opposite ends of two perpendicular diameters of the ring. The new angular velocity of the ring will be

MωM+4m
(M+4m)ωM
(M−4m)ωM+4m
Mω4m

Discuss Question

Answer: Option A. -> MωM+4m
:
A
Initial angular momentum of ring

= I ω = M R^{2} ω

If four object each of mass m, and kept gently to the opposite ends of two perpendicular
diameters of the ring then final angular momentum

= (M R^{2} + 4 m R^{2}) ω^{'}

By the conservation of angular momentum
Initial angular momentum = Final angular momentum

M R^{2} ω = (M R^{2} + 4 m R^{2}) ω^{'} \Rightarrow ω^{'} = (\frac{M}{M + 4 m}) ω .

Question 16. A sphere rolls down on an inclined plane of inclination

θ

. What is the acceleration as the sphere reaches bottom

57gsinθ
35gsinθ
27gsinθ
25gsinθ

Discuss Question

Answer: Option A. -> 57gsinθ
:
A
Acceleration (a)

= \frac{g sin θ}{1 + \frac{K^{2}}{R^{2}}} = \frac{g sin θ}{1 + \frac{2}{5}} = \frac{5}{7} g sin θ .

Question 17. A solid sphere rolls down an inclined plane and its velocity at the bottom is

v_{1}

. Then same sphere slides down the plane (without friction) and let its velocity at the bottom be

v_{2}

. Which of the following relation is correct

v1=v2
v1=57v2
v1=75v2
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
When solid sphere rolls down an inclined plane the velocity at bottom

v_{1} = \sqrt{\frac{10}{7} g h}

but, if there is no friction then it slides on inclined plane and the velocity at bottom

v_{2} = \sqrt{2 g h}

∴ \frac{v_{1}}{v_{2}} = \sqrt{\frac{5}{7}} .

Question 18. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its center. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity

ω_{0}

. When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform

ω

(t) will vary with time t as

Discuss Question

Answer: Option D. -> None of these
:
B
The angular momentum (L) of the system is conserved i.e. L =

I ω

= constant
When the tortoise walks along a chord, it first moves closer to the centre and then away from the centre. Hence, M.I. first decreases and then increases. As a result,

ω

will firstincrease and then decrease. Also the change in

ω

will be non-linear function of time.

Question 19. A ring, a solid sphere and a thin disc of different masses rotate with the same kinetic energy. Equal torques are applied to stop them. Which will make the least number of rotations before coming to rest

Disc
Ring
Solid sphere
All will make same number of rotations

Discuss Question

Answer: Option D. -> All will make same number of rotations
:
D
As

W = τ θ = E n e r g y \Rightarrow θ = \frac{E n e r g y}{τ} = 2 n π

So, if energy and torque are same then all the bodies will make same number of rotation.

Question 20. A metre stick is held vertically with one end on a rough horizontal floor. It is gently allowed to fall on the floor. Assuming that the end at the floor does not slip, find the angular speed of the rod when it hits the floor