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Examples of circular motion - class-XI

Description: examples of circular motion
Number of Questions: 20
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Tags: physics circular motion and gravitation uniform circular motion
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The tangential acceleration at particle at the lowest position is 

  1. 0

  2. 2 $m/{ s }^{ 2 }$

  3. 1 $m/{ s }^{ 2 }$

  4. None of these


Correct Option: C

A small bucket containing water is rotated in a vertical circle of radius $R$ by means of a rope. $v$ is velocity of bucket at highest point. Then water does not fall down if:

  1. $v \leq \sqrt{gR}$

  2. $v \leq \sqrt{\frac{gR}{2}}$

  3. $v \geq \sqrt{gR}$

  4. $v \geq \sqrt{\frac{gR}{2}}$


Correct Option: C
Explanation:

At highest point the centripetal acceleration should be greater than g.
$\therefore \dfrac{v^{2}}{R}\geq g$


or, $v\geq \sqrt{Rg}$

A bucket filled with water is tied to a rope of length $0.5\ m$ and is rotated in a circular path in vertical plane. the least velocity it should have at the lowest point of circle so that water does not spill is $(g=10ms^{-2})$:

  1. $\sqrt{5}\ m/s$

  2. $\sqrt{10}\ m/s$

  3. $5\ m/ s$

  4. $2\sqrt{5}\ m/s$


Correct Option: C
Explanation:

Minimum velocity at bottom $V=\sqrt{5g\ell}$ $=\sqrt{5\times 10\times 0.5}$$=5:m/s$


A very small particle rests on the top of a hemisphere of radius $20\ cm$. The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface taking $g = 9.8 m/s^{2}$ is:

  1. $\sqrt{9.8}\ m/s$

  2. $\sqrt{4.9}\ m/s$

  3. $\sqrt{1.96}\ m/s$

  4. $\sqrt{3.92}\ m/s$


Correct Option: C
Explanation:

Critical velocity at the top most point is $\sqrt{g\ell}$
$\therefore$ smallest velocity, such that the particle just leaves the surface =$\sqrt{9.8\times 0.2}=\sqrt{1.96}=1.4\ ms^{-1}$

The minimum centripetal force required to rotate a body mass $m$ in a vertical circle of radius $r$ is

  1. $mg$

  2. $2mg$

  3. $\dfrac{mg}{2}$

  4. Zero


Correct Option: A
Explanation:

The tension at the top-most point would be zero in the critical case that the particle just completes the verticle circle without being slake. The minimum centripetal force would then be equal to the force due to gravity $=mg$


A particle of mass m is rotating by means of a string in a vertical circle. The difference in the tensions at the bottom and the top towards completion of a full revolution would be

  1. 6 mg

  2. 4 mg

  3. 5 mg

  4. 2 mg


Correct Option: A
Explanation:

The tension at any point on the circular path is given by $T=(m/R)(u^2+gr-3gh)$; h is the height of the point from the horizontal

For a full circle, $h =2R$ Substituting, we get, $T _H=(m/R)(u^2-5gR)$ and 
at the bottom point of the circle , h=0 and thus, $T _L=(m/R)(u^2+gR)$

The difference in tension will be  $(T _L-T _H)=(m/R)(6gR) = 6mg$

Thus, the correct option is (a)

Four objects of masses m, 2m , 3m and 4m are attached to a string of length L=2 metres and rotated vertically with a speed of $\sqrt(5Lg)$, pick out the most appropriate one:

  1. m

  2. 2m

  3. 4m

  4. All the objects will complete the circle


Correct Option: D
Explanation:

For an object to complete one full circle, the initial velocity should be atleast $\sqrt{5Lg}$. Thus, it is independent of the mass used

The correct option is (d)

An object suspended by a string has a critical velocity at the top most point of a circle. It will complete a full circle 

  1. True

  2. False


Correct Option: A
Explanation:

Critical velocity is the minimum velocity at the top most point of a circle in which it is said to complete one full revolution. 

Thus, the statement is true

A block of mass $m$ at the end of a string is whirled round in a vertical circle of radius $R$. The critical speed of the block at the top of its swing below which the string would slacken before the block reaches the top is:

  1. $Rg$

  2. ${(Rg)}^{2}$

  3. $R/g$

  4. $\sqrt {Rg}$


Correct Option: D
Explanation:

At the top if the block is moving with critical speed the tension in the string would be zero and therefore force of gravity would provide the necessary centripetal force:

$mg=\cfrac { m{ v }^{ 2 } }{ R } \ \Rightarrow v=\sqrt { Rg } $

A bob of mass $100\ g$ tied at the end of a string of length $50\ cm$ is revolved in a vertical circle with constant speed of $1\ ms^{-1}$. When the tension in the string is $0.7N$, the angle made by the vertical is $(g=10\ ms^{-2})$

  1. $0^{o}$

  2. $90^{o}$

  3. $180^{o}$

  4. $60^{o}$


Correct Option: C

A weightless thread can withstand tension upto 30 N.A. stone of mass 0.5kg is tied to it and is revolved in a circular path of radius 2m i n a vertical plane. If $g=10m/s^2$, then the maximum angular velocity of the stone can be;-

  1. $5\, rad / s$

  2. $\sqrt{30} \ rad/s$

  3. $\sqrt{60} \ rad/s$

  4. $10\, rad / s$


Correct Option: A

An inclined plane ends into vertical loop of radius $R$. A particle is released from height $3R$. Can it loop the loop?

  1. yes

  2. no

  3. cannot say

  4. yes if fraction is present


Correct Option: A
Explanation:

The minimum speed required at the bottom of the circle(r) to complete the circular motion is $\sqrt {5gR}$
For this speed, the minimum height required is
$0.5mv^2=mgh$
$\Rightarrow 0.5(5gR)=gh$
$\Rightarrow h=\dfrac {5}{2}R$
So, the minimum height from which the body has to released is 2.5R. We have a height of 3R, so the particle completes the vertical loop.
Option A.

A block of mass $m$ at the end of a string is whirled round in a vertical circle of radius $r$. The critical speed of the block at the top of its swing below which the string would slacken before block reaches the top is

  1. $\sqrt{2rg}$

  2. $\sqrt{3rg}$

  3. $\sqrt{rg}$

  4. $\sqrt{5rg}$


Correct Option: C
Explanation:

the string will not slack if the centrifugal force on the particle is greater then or equal to the weight of mass i.e.
$\dfrac{mv^2}{r} \ge mg \ \Rightarrow v \ge \sqrt{rg} \ \Rightarrow v _c=\sqrt{rg} $

The maximum tension that an inextensible ring of radius 1 m and mass density 0.1 kg ${ m }^{ -1 }$ can bear is 40 N . The maximum angular velocity with which it can be rotated in a  circular path is 

  1. 20 rad/s

  2. 18 rad/s

  3. 16 rad/s

  4. 15 rad/s


Correct Option: A

A body is allowed to slide on a frictionless track from rest under-gravity.The track ends in a circular of diameter D. What should be the mini-mum height of the body in terms of D.          So that it may successfully complete the loop?

  1. $\dfrac{9}{4}D $

  2. $ \frac {5}{4} D $

  3. D

  4. 2 D


Correct Option: A
Explanation:

Velocity at top of the circle is $\sqrt{5gr}$ 

Energy lost to reach height D from height H  $ = mg(H-D)$

Energy gained in the form of $K.E= \dfrac12 mv^2= \dfrac12 m(5gr) = \dfrac{5mgD}{4}$ 

Equating K.E. and P.E. ,

$\dfrac{5mgD}{4} = mg(H-D)$

$H=\dfrac{9D}{4}$

A body is a allowed to slide down a frictionless track from rest position at its top under gravity. The track ends in a circular loop of diameter $D$. Then, the minimum height of the inclined track  (in terms of $D$ ) so that it may complete successfully the loop is:

  1. $\dfrac{7D}{4}$

  2. $\dfrac{9D}{4}$

  3. $\dfrac{5D}{4}$

  4. $\dfrac{3D}{4}$


Correct Option: C
Explanation:

$V _{0}$ required for a complete vertical circle: $V _{0} =\sqrt{\dfrac{5}{2}gD}$
WET between top and bottom:
              $mgH  = \dfrac{1}{2}m V _{0}^{2}$


                 $gH  = \dfrac{5}{4}gD$

                   $H  = \dfrac{5}{4}D$

The length of simple pendulum is $1m$ and mass of its bob is $50 gram$. The bob is given  sufficient velocity so that the bob describes vertical circle whose radius equal to length of pendulum, the tension in the string at lowest extreme position is:

  1. $2.5 N$

  2. $1N$

  3. $1.5 N$

  4. $2N$


Correct Option: A
Explanation:

Using conservation of energy, 

$mg \times 2R = mv^2/2$
$v=2 \sqrt{Rg}$ .........(1)

At lower most point, from free body digram,
$T - mg = m v^2/R$ ...........(2)
From (1) and (2), 
$T = 5mg$
$T = 2.5N$

A weightless thread can bear tension up to $3.7kg-wt$. A stone of mass $500g$ is tied to it and revolved in a circular path of radius $4$m in a verticle plane. If $g=10m/s^2$, then the maximum angular velocity of the stone will be:

  1. $2 rad/s$

  2. $4 rad/s$

  3. $16 rad/s$

  4. $\sqrt{21} rad/s$


Correct Option: B
Explanation:

Tension is maximum at lowest point of a verticle circle $T _{max}=mr\omega^2+mg$
$3.7g=0.5\times 4\times \omega^2+0.5g$
$2\omega^2=3.2g$
$\omega=\sqrt{1.6g}=\sqrt{1.6\times 10}$
$\omega=4rad/s$

A body attached to a string of length describes a vertical circle such that it is just able to cross the highest point. Find the minimum velocity at the bottom of the circle.

  1. $\sqrt { 2gL } $

  2. $\sqrt { gL } $

  3. $\sqrt { 3gL } $

  4. $\sqrt { 5gL } $


Correct Option: B
Explanation:
Radius of the circle traced will be equal to the length of the string
In circular motion tension in the string is balanced by the centripetal force
hence we know that,
$T=\dfrac { m{ v }^{ 2 } }{ r } $ 
$T=\dfrac { m{ v }^{ 2 } }{ L } $
Velocity at any point,$V=\sqrt { \dfrac { TL }{ m }  } $
And at lowest point the tension is balanced by the weight of the block
 tension = weight = mg
$V=\sqrt { \dfrac { mgL }{ m }  } =\sqrt { gL } $

A bucket is whirled in a vertical circle with a string attached to it. The water in the bucket does not fall down even when the bucket is inverted at the top of its path. We can say that in this position.

  1. $mg=\displaystyle\frac{mv^2}{r}$

  2. mg is greater than $\displaystyle\frac{mv^2}{r}$

  3. mg is not greater than $\displaystyle\frac{mv^2}{r}$

  4. mg is not less than $\displaystyle\frac{mv^2}{r}$


Correct Option: A,C
Explanation:

When the bucket is at the top of its path, the forces acting on the water are gravity and centripetal force. Since, the water does not fall down. Therefore, centripetal force is greater than the force of gravity.

Therefore, $\cfrac{mV^2}{R}\ge mg$

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