Rotation Rock And Roll(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

ROTATION ROCK AND ROLL MCQs

Total Questions : 30 | Page 1 of 3 pages

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

$\frac{M ω}{M + 4 m}$
$\frac{(M + 4 m) ω}{M}$
$\frac{(M - 4 m) ω}{M + 4 m}$
$\frac{M ω}{4 m}$

Discuss Question

Answer: Option A. ->

\frac{M ω}{M + 4 m}

:
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 2.

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 B. ->

:
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 first increase and then decrease. Also the change in

ω

will be non-linear function of time.

Question 3.

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

$\frac{I_{1} ω_{1} + I_{2} ω_{2}}{2 (I_{1} + I_{2})}$
$\frac{(I_{1} ω_{1} + I_{2} ω_{2})^{2}}{2 (I_{1} + I_{2})}$
$\frac{(I_{1} + I_{2}) (ω_{1} + ω_{2})^{2}}{2}$
None of these

Discuss Question

Answer: Option B. ->

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

:
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 4.

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 5.

A solid cylinder of mass 'M' and radius 'R' rolls without slipping down an inclined plane of length 'L' and height 'h'. What is the speed of its centre of mass when the cylinder reaches its bottom

$\sqrt{\frac{3}{4} g h}$
$\sqrt{\frac{4}{3} g h}$
$\sqrt{4 g h}$
$\sqrt{2 g h}$

Discuss Question

Answer: Option B. ->

\sqrt{\frac{4}{3} g h}

:
B
Velocity at the bottom (v)

= \sqrt{\frac{2 g h}{1 + \frac{K^{2}}{R^{2}}}} = \sqrt{\frac{2 g h}{1 + \frac{1}{2}}} = \sqrt{\frac{4}{3} g h} .

Question 6.

A sphere rolls down on an inclined plane of inclination $θ$ . What is the acceleration as the sphere reaches bottom

$\frac{5}{7} g sin θ$
$\frac{3}{5} g sin θ$
$\frac{2}{7} g sin θ$
$\frac{2}{5} g sin θ$

Discuss Question

Answer: Option A. ->

\frac{5}{7} g sin θ

:
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 7.

A thin uniform circular ring is rolling down an inclined plane of inclination $30^{\circ}$ without slipping. Its linear acceleration along the inclined plane will be

$\frac{g}{2}$
$\frac{g}{3}$
$\frac{g}{4}$
$\frac{g}{5}$

Discuss Question

Answer: Option C. ->

\frac{g}{4}

:
C

a = \frac{g sin θ}{1 + \frac{k^{2}}{R^{2}}} = \frac{g sin 30^{\circ}}{1 + 1} = \frac{g}{4}

[A s \frac{k^{2}}{R^{2}} = 1 a n d θ = 30^{\circ}]

Question 8.

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

$\frac{2 r}{3}$
$\frac{2 r}{5}$
$\frac{r}{2}$
$\frac{3 r}{4}$

Discuss Question

Answer: Option A. ->

\frac{2 r}{3}

:
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 9.

A small meteorite of mass 'm' travelling towards the centre of earth strikes the earth at the equator. The earth is a uniform sphere of mass 'M' and radius 'R'. The length of the day was 'T' before the meteorite struck. After the meteorite strikes the earth, the length of day increases (in sec) by

$\frac{5 m T}{2 M}$
$\frac{m}{M} T$
$\frac{4 m T}{5 M}$
$\frac{M}{3 m T}$

Discuss Question

Answer: Option A. ->

\frac{5 m T}{2 M}

:
A

I_{1} ω_{1} = I_{2} ω_{2}

\frac{2}{5} M R^{2} \times (\frac{2 π}{T_{1}}) = (\frac{2}{5} M R^{2} + m R^{2}) \frac{2 π}{T_{2}}

\frac{T_{2}}{T_{1}} = 1 + \frac{5 m}{2 M}

\frac{T_{2} - T_{1}}{T_{1}} = \frac{5 m}{2 M}

T_{2} - T_{1} = \frac{5 m T}{2 M}

[i . e; T_{1} = T]

Question 10.

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