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 1 of 3 pages

Question 1. A small body slides over the curved surface of a semicircular fixed cylinder of radius 'r', kept horizontally on the ground as shown in the figure. At what height from the ground would the body lose contact with the surface

Discuss Question

Answer: Option A. -> 2r3
:
A
Let normal reaction makes an angle

θ

from vertical, then

v^{2} = 2 g r (1 - cos θ)

and

\frac{m v^{2}}{r} = m g cos θ

\Rightarrow

height from ground

h = \frac{2 r}{3}

Question 2. A uniform rod of mass 'm' and length 'l' is kept vertical with the lower end clamped. It is slightly pushed to let it fall down under gravity. Find its angular speed when the rod is passing through its lowest position. Neglect any friction at the clamp. What will be the linear speed of the free end at this instant?

√6gl,√6gl
√2gl,√6gl
√6gl,√3gl
√3gl,√6gl

Discuss Question

Answer: Option A. -> √6gl,√6gl
:
A

A Uniform Rod Of Mass 'm' And Length 'l' Is Kept Vertical Wi...

As the rod reaches its lowest position, the centre of mass is lowered by a distance l. Its gravitational potential energy is decreased by mgl. As no energy is lost against friction, this should be equal to the increase in the kinetic energy. As the rotation occurs about the horizontal axis through the clamped end, the moment of inertia is I =

m l^{2} / 3

. Thus,

\frac{1}{2} I ω^{2} = m g l

\frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} = m g l

ω = \sqrt{\frac{6 g}{l}}

.
The linear speed of the free end is

V = I ω = \sqrt{6 g l}

Question 3. 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 4. A metal ball of mass 'm' is put at the point A of a loop track and the vertical distance of A from the lower most point of track is 8 times the radius 'R' of the circular part. The linear velocity of ball when it rolls of the point B to a height 'R' in the circular track will be

[10gR]1/2
7[gR10]1/2
[7gR5]1/2
[5gR]1/2

Discuss Question

Answer: Option A. -> [10gR]1/2
:
A
Applying the conservation of energy at points A and B, we have

m g (8 R) = \frac{m v^{2}}{2} + \frac{1}{2} I ω^{2} + m g R

m g (8 R) = \frac{m v^{2}}{2} + \frac{1}{2} (\frac{2}{5} m R^{2}) {(\frac{v}{R})}^{2} + m g R

= \frac{7}{10} m v^{2} + m g R

m g (8 R - R) = \frac{7}{10} m v^{2}

[10 g R]^{1 / 2}

Question 5. The torque

⃗ τ

on a body about a given point is found to be

⃗ A \times ⃗ L

where

⃗ A

is a constant vector and

⃗ L

is angular momentum of the body about that point. From this it follows that

d⃗Ldt is perpendicular to ⃗L at all times.
the component of ⃗L in the direction of ⃗A does change with time.
the magnitude of ⃗L does change with time.
⃗L does not change with time.

Discuss Question

Answer: Option A. -> d⃗Ldt is perpendicular to ⃗L at all times.
:
A
Due to law of conservation of angular momentum,

⃗ L

= constant
i.e.

⃗ L . ⃗ L

= constant
or,

\frac{d}{d t} (⃗ L . ⃗ L) = 0

or,

2 ⃗ L . \frac{d ⃗ L}{d t} = 0

or,

⃗ L ⊥ \frac{d ⃗ L}{d t}

Since

τ = ⃗ A \times ⃗ L

\frac{d ⃗ L}{d t} = ⃗ A \times ⃗ L

i.e.,

\frac{d ⃗ L}{d t}

must be perpendicular to

⃗ A

as well as

⃗ L

.
Further the component of

⃗ L

along

⃗ A

\frac{⃗ A . ⃗ L}{A} .

Also

\frac{d}{d t} (⃗ A . ⃗ L) = ⃗ A . \frac{d ⃗ L}{d t} + ⃗ L . \frac{d ⃗ A}{d t} = 0

{∵ ⃗ A ⊥ \frac{d ⃗ L}{d t} a n d \frac{d ⃗ A}{d t} = ⃗ 0}

or,

⃗ A . ⃗ L

= constant
i.e.,

\frac{⃗ A . ⃗ L}{A}

= x = constant
Since

\frac{d ⃗ L}{d t} (o r τ)

is perpendicular to

⃗ L

, hence it cannot change magnitude of

⃗ L

but can surely change direction of

⃗ L

Question 6. A smooth sphere A is moving on a frictionless horizontal plane with angular speed

ω

and centre of mass is moving translationally with velocity 'v'. It collides elastically and head-on with an identical sphere B which is at rest. All surfaces are frictionless. After the collision, their angular speeds are

ω_{A}

and

ω_{B}

, respectively. Then,

ωB=0
ωA=ωB
ωA
ωB=ω

Discuss Question

Answer: Option A. -> ωB=0
:
A

ω_{B} = 0

rotational KE can’t be transferred from A to B as surfaces are frictionless.

Question 7. A smooth uniform rod of length 'L' and mass 'M' has two identical beads of negligible size, each of mass 'm', which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with angular velocity

ω_{0}

about an axis perpendicular to the rod and passing through the mid point of the rod (see figure). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is

A Smooth Uniform Rod Of Length 'L' And Mass 'M' Has Two Iden...

ω0
Mω0M+12m
Mω0M+2m
Mω0M+6m

Discuss Question

Answer: Option D. -> Mω0M+6m
:
D
Since there are no external forces therefore the angular momentum of the system remains constant.
Initially when the beads are at the centre of the rod angular momentum

L_{1} = (\frac{M L^{2}}{12}) ω_{0}

.....(i)
When beads reach the ends of the rod then angular momentum

= (m {(\frac{L}{2})}^{2} + m {(\frac{L}{2})}^{2} + \frac{M L^{2}}{12}) ω^{'}

..(ii) Equating (i) and (ii)

\frac{M L^{2}}{12} ω_{0} = (\frac{m L^{2}}{2} + \frac{M L^{2}}{12}) ω^{'} \Rightarrow ω^{'} = \frac{M ω_{0}}{M + 6 m} .

Question 8. A solid sphere of mass 0.1 kg and radius 2 cm rolls down an inclined plane 1.4m in length at an angle whose slope is 1/10. If sphere started from rest, its final velocity will be

1.4 m/sec
0.14 m/sec
14 m/sec
0.7 m/sec

Discuss Question

Answer: Option A. -> 1.4 m/sec
:
A

v = \sqrt{\frac{2 g h}{1 + \frac{k^{2}}{R^{2}}}} = \sqrt{\frac{2 \times 9.8 \times l sin θ}{1 + \frac{2}{5}}} [A s \frac{k^{2}}{R^{2}} = \frac{2}{5}, l = \frac{h}{s i n θ} a n d sin θ = \frac{1}{10} g i v e n]

\Rightarrow v = \sqrt{\frac{2 \times 9.8 \times 1.4 \times \frac{1}{10}}{\frac{7}{5}}} = 1.4 m / s .

Question 9. In the above problem the angular velocity of the system after the particle sticks to it will be

0.3 rad/s
5.3 rad/s
10.3 rad/s
89.3 rad/s

Discuss Question

Answer: Option C. -> 10.3 rad/s
:
C
Initial angular momentum of bullet + initial angular momentum of cylinder
= Final angular momentum of (bullet + cylinder) system

\Rightarrow m v r + I_{1} ω = (I_{1} + I_{2}) ω^{'}

\Rightarrow m v r + I_{1} ω = (\frac{1}{2} M r^{2} + m r^{2}) ω^{'}

\Rightarrow 0.5 \times 5 \times 0.2 + 0.12 = (\frac{1}{2} 2 (0.2)^{2} + (0.5) (0.2)^{2}) ω^{'}

∴ ω^{'} = 10.3 r a d / s e c .

Question 10. A cord is wound round the circumference of wheel of radius 'r'. The axis of the wheel is horizontal and moment of inertia about it is 'I'. A weight 'mg' is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be