Tag: equation of motion of rotating body

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Questions Related to equation of motion of rotating body

Multiple choice 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

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 :

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

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

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

  4. $\cfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice 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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $W= \tau \theta$

Multiple choice 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

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

  1. $\tau=dL/dt$

  2. $\tau=dp/dt$

  3. $\tau=mdv/dt$

  4. $\tau=md\alpha/dt$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Newton's second law for rotation states that the net torque acting on an object is equal to the rate of change of its angular momentum, expressed as tau = dL/dt.

Multiple choice 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

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:

  1. $m \omega^2/2$

  2. $I \omega^2/2$

  3. $m V^2/2$

  4. $I V^2/2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)

Multiple choice 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

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

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

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

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

  4. $\dfrac { 2 }{ 5 } g\sin \theta$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)
$a = \alpha R$
$\therefore a = \dfrac{59}{7} \sin \theta$

Multiple choice 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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For half a revolution, the center moves forward by pi*R. The point initially at the bottom moves to the top, so its vertical displacement is 2R. The total displacement is sqrt((pi*R)^2 + (2R)^2) = R*sqrt(pi^2 + 4). The angle is tan^-1(vertical/horizontal) = tan^-1(2R/pi*R) = tan^-1(2/pi).

Multiple choice 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

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

  1. Decrease its mechanical kinetic energy

  2. Increase its translational energy

  3. Increases its rotational kinetic energy

  4. Decreases its potential energy
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As a cylinder rolls down an incline, the static friction force acts up the plane. This torque causes the object to rotate, thereby increasing its rotational kinetic energy at the expense of potential energy.

Multiple choice 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

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 = 

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

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

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

  4. $\frac{2UH^2}{gW^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ball hits the nth step when its vertical displacement is nH and horizontal displacement is nW. Using y = (1/2)gt^2 and x = Ut, we get nH = (1/2)g(nW/U)^2. Solving for n gives n = 2U^2H / (gW^2).

Multiple choice 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

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. 

  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. $\left( \cfrac { kQqR }{ mg } \right) ^{ 3 }$
Reveal answer Fill a bubble to check yourself
D Correct answer