Physics

Rotational and Circular Motion

235 Questions

Rotational and circular motion examines the dynamics of objects moving in circular paths or rotating around an axis. Key concepts include angular momentum, torque, moment of inertia, and centripetal force. This is a highly scoring topic in the physics section of competitive exams.

Angular momentumCentripetal forceMoment of inertiaRolling objectsGyroscopic effect

Rotational and Circular Motion Questions

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

A thin uniform rod of length l and m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of:

  1. $\dfrac{1}{6} \dfrac{l\omega}{g}$
  2. $\dfrac{1}{2} \dfrac{l^2\omega^2}{g}$
  3. $\dfrac{1}{6} \dfrac{l^2\omega^2}{g}$
  4. $\dfrac{1}{3} \dfrac{l^2\omega^2}{g}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

At lowest height

$E=\cfrac{1}{2}I \omega^2$
At maximum weight $\omega=0$
$E=mgh+0$ (his height from lower point)
By conservation of energy
$\cfrac{1}{2}I\omega^2=mgh\h=\cfrac{I\omega^2}{2mg}\h=\cfrac{ml^2\omega^2}{6mg}=\cfrac{l^2\omega^2}{6g}$

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

A straight rod of length L has one of its ends at the origin and the other at $x=L$. If the mass per unit length of the rod is given by Ax where A is constant, where is its mass centre?

  1. $L/3$
  2. $L/2$
  3. $2L/3$
  4. $3L/4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
I assume you meant to say "a is a CONSTANT".

xc = coordinate of center of mass

M = total mass

$xc = ∫xdm / ∫dm = ∫xdm / M$

Given:$ m(x) = ax ⇒ dm/dx = a ⇒ dm = adx$

$∫xdm = ∫01 x(adx) = a/2$

$M = ∫dm = ∫(dm/dx)dx = ∫01 adx = a$

$xc = (a/2)/a = 1/2$

By the way, this is a mechanics problem (in statics), not a thermodynamics problem.

 

Multiple choice physics energy : forms and sources concept of energy energy for everything forms of energy

A particle of mass $m$ is moving in a horizontal circle of radius $r$ with a centripetal force $\vec {F}=\dfrac{-k}{r^{2}}\hat {r}$. The total mechanical energy of the particle is 

  1. $\dfrac{-k}{r}$
  2. $\dfrac{3}{2}\dfrac{k}{r}$
  3. $\dfrac{-k}{2r}$
  4. $\dfrac{-3k}{2r}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a central force F = -k/r^2, the potential energy U = -k/r. The kinetic energy K = 1/2 * |U| = k/2r. Total energy E = K + U = k/2r - k/r = -k/2r.

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

The rod of fixed length $k$ slides along the coordinate axes. If it meets the axes at $A(a,0)$ and $B(0,b)$, then the minimum value of $\left(a+\dfrac 1a\right) ^2+\left(b+\dfrac 1b\right) ^2$ is

  1. $0$
  2. $8$
  3. $k^2-4+\dfrac 4{k^2}$
  4. $k^2+4+\dfrac 4{k^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

since, $A.M. \geq G.M. \implies \dfrac{a+b}{2}\geq \sqrt{ab}$


let $a=1 $ and $b=a^2$

Therefore, $\dfrac{1+a^2}{2}\geq \sqrt{a^2}$

$\implies \dfrac{1+a^2}{2}\geq \sqrt{a^2}$

$\implies \dfrac{1+a^2}{2}\geq a$

$\implies \dfrac{1+a^2}{a}\geq 2$

$\implies \dfrac{1}{a}+\dfrac{a^2}{a}\geq 2$

$\implies a+\dfrac{1}{a}\geq 2$
 squaring on both sides
$\implies (a+\dfrac{1}{a})^2\geq 4$ ------- (1)

similarly, $ (b+\dfrac{1}{b})^2\geq 4$ --------(2)

adding (1) and (2) we get

$(a+\dfrac{1}{a})^2+ (b+\dfrac{1}{b})^2\geq 4+4$

$(a+\dfrac{1}{a})^2+ (b+\dfrac{1}{b})^2\geq 8$

Therefore the minimum value is $8$

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

The angular frequency of a torsional pendulum is $\omega$ rad/s. If the moment of inertia of the object is I, the torsional constant of the wire is related to the rotational kinetic energy of the disc, if the disc was rotating with an angular velocity $\omega$ is

  1. k= 2 KE

  2. k= KE

  3. k= 4 KE

  4. k= KE/2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

we know that $T=2 \pi \sqrt{I/k}$. Substituting the values given, we get, $k= I \omega^2 =2 \times $ kinetic energy

The correct option is (a)

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are
attached at distance $'L/2'$ from its centre on both sides, it reduces the oscillation frequency by $20\%$. The value of ratio m/M is close to :

  1. 0.175

  2. 0.375

  3. 0.575

  4. 0.775
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Frequency of torsonal oscillations is given by 
$f = \dfrac{k}{\sqrt{I}}$
$f _1 = \dfrac{k}{\sqrt{\dfrac{M (2L)^2}{12}}}$
$f _2 =  \dfrac{k}{\sqrt{\dfrac{M (2L)^2}{12} + 2m \left(\dfrac{L}{2} \right)^2}}$
$f _2 = 0.8 f _1$
$\dfrac{m}{M} = 0.375$

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A metallic disc oscillates about an axis through its edge in it's own plane. The equivalent length of the disc as a pendulum is

  1. $r$
  2. $\dfrac r3$
  3. $\dfrac { r } { 2 }$
  4. $\dfrac { 3r } { 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a disc pivoted at its edge, I = I_cm + mR^2 = (1/2)mR^2 + mR^2 = (3/2)mR^2. The equivalent length L_eq = I/(mR) = (3/2)R.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A disc of masses m and radius 2r are suspended through a fine wire of torsional constant K. The wire is attached to the centre of the plane of the disc and given torsional oscillations. If the disc is replaced by another disc of mass 4m and radius 2r, the ratio of the time period of oscillations are

  1. 4:1

  2. 1:4

  3. 1:1

  4. 2:1
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics measurement of physical quantities kinds of units fundamental quantities standardized measurement

The impulse $J$ required to bring it to rest when it reaches the vertical position.

  1. Impulse of $m\sqrt {\dfrac {gl}{3}}$ when it is applied horizontally at distance $l$ from pivoted point
  2. Impulse of $m\sqrt {\dfrac {gl}{3}}$ when it is applied horizontally at distance $\dfrac {l}{2}$ from pivoted point
  3. Minimum impulse is $m\sqrt {\dfrac {gl}{2}}$ when it is applied at distance $l$ from pivoted point
  4. If rod is stopped by applying minimum impulse then the is impulse imparted by pivot is $\dfrac {m\sqrt {gl}}{2\sqrt {3}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

The following data were obtained in experiment on inversion of cane sugar.
Time (minutes)         0        60        120      180      360     $\infty $
Angle of rotation  +13.1   +11.6   +10.2   +9.0   +5.87  -3.8
   (degree)
Determine total time ?

  1. 966 min

  2. 483 min

  3. 1932 min

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The integrated rate law expression for the inversion of can sugar (assuming first order kinetics) is as shown.
$\displaystyle k = \frac {2.303}{t} log \frac {r _0 - r _{\infty}}{r _t - r _{\infty}} $
For 60 minutes
$\displaystyle k = \frac {2.303}{60} log \frac {13.1 - (-3.8)}{11.6 - (3.8)} = 0.001549 $
For 120 minutes
$\displaystyle k = \frac {2.303}{120} log \frac {13.1 - (-3.8)}{10.2 - (3.8)} = 0.001569 $
For 180 minutes
$\displaystyle k = \frac {2.303}{180} log \frac {13.1 - (-3.8)}{9.0 - (3.8)} = 0.001544 $
For 360 minutes
$\displaystyle k = \frac {2.303}{360} log \frac {13.1 - (-3.8)}{5.87 - (3.8)} = 0.001551 $
Since, the value of k is constant, the reaction follows first order reaction.
The average value of k is $\displaystyle  \frac {0.001549+0.001569+0.001544+0.001551}{4} = \frac {0.0062135}{4} = 0.001553 : min^{-1}$
To determine the total time, substitute $\displaystyle r _t = 0 $ in the above expression.
$\displaystyle t = \frac {2.303}{k} log \frac {r _0 - r _{\infty}}{r _t - r _{\infty}} $
$\displaystyle t = \frac {2.303}{0.001553} log \frac {13.1 - (-3.8)}{0 - (-3.8)} = 966 : min $

Multiple choice physics the essence of change forms of energy and energy conservation energy for everything forms of energy

A small sphere is moving at a constant speed in a vertical circle. Below is a list of quantities that could be used to describe some aspect of the motion of the sphere.
$I$   -kinetic energy
$II$  -gravitational potential energy
$III$  - momentum
Which of these quantities will change as this sphere moves around the circle?

  1. $I$ and $II$ only
  2. $I$ and $III$ only
  3. $III$ only
  4. $II$ and $III$ only
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Speed is given as constant. Though the velocity is changing only in direction, its magnitude remains the same. Thus the kinetic energy remains constant as it is a scalar quantity but momentum would change as it is a vector quantity.
Also as the elevation of the ball keeps on changing during the motion, the potential energy changes

Multiple choice modelling collisions collisions momentum work, energy and power physics

A solid cylinder of mass 'M' and radius 'R' is rotating along its axis with angular velocity $\omega $ without friction. A particle of mass 'm' moving with velocity v collide against the cylinder and sticks to its rim. After the impact calculate angular velocity of cylinder.

  1. $\cfrac { I+R\omega }{ I+m{ R }^{ 2 } } $
  2. $\cfrac { mvR+IR }{ I+m{ R }^{ 2 } } $
  3. $\cfrac { I\omega +mvR }{ I+m{ R }^{ 2 } }$
  4. $\cfrac { I\omega +mR }{ I+mv{ R }^{ 2 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using conservation of angular momentum about the axis of the cylinder: L_initial = I*omega + m*v*R. L_final = (I + m*R^2)*omega_final. Equating them gives omega_final = (I*omega + m*v*R) / (I + m*R^2).