Tag: constant angular acceleration

Questions Related to constant angular acceleration

A solid sphere of mass 0.5 kg and diameter 1 m rolls without sliding with a constant velocity of 5 m/s, the ratio of the rotational K.E. to the total kinetic energy of the sphere is :

physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

  1. $\cfrac{7}{10}$

  2. $\cfrac{4}{9}$

  3. $\cfrac{2}{7}$

  4. $\cfrac{1}{2}$

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Correct Option: B

Analogue of mass in rotational motion is

physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

  1. Moment of inertia

  2. Angular momentum

  3. Gyration

  4. None of these

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Correct Option: A
Explanation:

Analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in translational motion.

Newton's second law of motion and work done in rotation of a rigid body can be expressed as 

physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

  1. Newton's law cannot be expressed in rotation, work done in rotation is $W=\tau \ theta$

  2. Force and work done are expressed as $\tau = I \alpha$ and $W=\tau \ theta$

  3. Force can be expressed as $\tau = I \alpha$, while work done will be zero

  4. Force will be zero, since no net displacement is present

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Correct Option: B
Explanation:

Newton's second law of motion in kinematics is F =  ma. To express the same in rotation, replace mass by moment of inertia I AND linear acceleration a by angular acceleration $\alpha$. Thus, Newton's law of motion becomes, Torque $\tau = I \alpha$

Similarly work done in kinematics is given by W = F.S. To express the same in rotation, replace Force by torque $\tau$ AND linear displacement by angular displacement $\theta$. Thus, Work done becomes, Torque $W= \tau \theta$

How do you express Newton's second law of motion in differential form

physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

  1. $\tau=dL/dt$

  2. $\tau=dp/dt$

  3. $\tau=mdv/dt$

  4. $\tau=md\alpha/dt$

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Correct Option: A

If Kinetic energy is expressed as $mv^2/2$ for a particle undergoing uniform velocity motion, How is the kinetic energy expressed in case of the same particle, if it was rotating:

physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

  1. $m \omega^2/2$

  2. $I \omega^2/2$

  3. $m V^2/2$

  4. $I V^2/2$

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Correct Option: B
Explanation:

In rotation, m is replaced by I and v by  $\omega$. Thus the expression for kinetic energy becomes $I \omega^2/2$

The correct option is thus option (b)

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

physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

  1. $\dfrac { 5 }{ 7 } g\sin \theta$

  2. $\dfrac { 3 }{ 5 } g\sin \theta$

  3. $\dfrac { 2 }{ 7 } g\sin \theta$

  4. $\dfrac { 2 }{ 5 } g\sin \theta$

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Correct Option: A
Explanation:

Net force = $mg \sin \theta$

Friction force = $F (\uparrow)$
For linear motion, 
$mg \, \sin \theta = f = mg$ ...(1)
angular motion 
$fR = I \alpha $ .... (2)
$\therefore mg \, \sin \theta = ma + \dfrac{I \alpha}{R}$ ....(3)
$a = \alpha R$
$\therefore a = \dfrac{59}{7} \sin \theta$

What is the displacement of the point on the wheel initially in contact with the ground when the wheel rolls forwards half of revolution? Take the radius of the wheel as $'R'$ and the x-axis in the forward direction

physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

  1. $R\sqrt{\pi^2 + 9}, Tan^{-1}\left(\dfrac{3}{\pi}\right)$ with x-axis

  2. $R\sqrt{\pi^2 + 4}$ and angle $Tan^{-1}\left(\dfrac{2}{\pi}\right)$ with x-axis

  3. $R\sqrt{\pi^2 + 16}, Tan^{-1}\left(\dfrac{4}{\pi}\right)$ with x-axis

  4. None

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Correct Option: B

A solid cylinder rolls down a rough inclined plane without slipping. As it goes down, what will happen due to force of friction?

physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

  1. Decrease its mechanical kinetic energy

  2. Increase its translational energy

  3. Increases its rotational kinetic energy

  4. Decreases its potential energy

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Correct Option: C

A ball rolling off the top of a staicase of each step with height H and width W, with an initial velocity U will just hit nth step. Then n = 

physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

  1. $\frac{2U^2H^2}{gW}$

  2. $\frac{2U^2H^2}{gW^2}$

  3. $\frac{2U^2H}{gW^2}$

  4. $\frac{2UH^2}{gW^2}$

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Correct Option: C

A small charged ball of mass m and charge q is suspended from the highers point of a ring of radius R by means of an insulated code of negligible mass.The ring is made of a rigid wire of negligible cross-section and lies in a vertical plane.On the ring, there is uniformly distributed charge Q of the same as that of q .determine the length of the cord so as the equilibrium position of the ball lies on the symmetry axis ,perpendicular to the plane of the ring. 

physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

  1. $\left( \cfrac { 2kQqR }{ mg } \right) ^{ 1/3 }$

  2. $\left( \cfrac { kQqR }{ mg } \right) ^{ 1/3 }$

  3. $\left( \cfrac { kQqR }{ 2mg } \right) ^{ 1/3 }$

  4. $\left( \cfrac { kQqR }{ mg } \right) ^{ 3 }$

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Correct Option: D