Tag: rotational motion of a rigid body and moment of inertia

Questions Related to rotational motion of a rigid body and moment of inertia

A uniform sphere is placed on a smooth horizontal surface and a horizontal force $F$ is applied on it at a distance $'h'$ above the surface. The acceleration of the centre

physics rotational motion of a rigid body and moment of inertia motion of rigid body rigid body equilibrium of a rigid body

  1. Is maximum when $h=0$

  2. Is maximum when $h=\dfrac { 2R }{ 5 }$

  3. Is maximum when $h=\dfrac { 7 }{ 5 } R$

  4. Is independent of $h$

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Correct Option: D

A block of mass $m$ slides down a smooth vertical circular track. During the motion, the block is in.

physics rotational motion of a rigid body and moment of inertia motion of rigid body rigid body equilibrium of a rigid body

  1. Vertical equilibrium

  2. Horizontal equilibrium

  3. Radial equilibrium

  4. None of these.

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Correct Option: A

A heavy seesaw (i.e., not massless) is out of balance. A light girl sits on the end that is tilted downward, and a heavy body sits on the other side so that the seesaw now balances. If they both move forward so that they are one-half their original distance from the pivot point (the fulcrum) what will happen to the seesaw?

physics rotational motion of a rigid body and moment of inertia motion of rigid body rigid body equilibrium of a rigid body

  1. The side the body is sitting on will tilt downward

  2. The side the girl is sitting on will once again tilt downward

  3. Nothing; the seesaw will still be balanced.

  4. It is impossible to say without knowing the masses and the distances.

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Correct Option: B
Explanation:

Let the mass distribution on the left of seesaw be x kg and that on the right be y kg such that $x>y$. And let the mass of the girl be $m _g$ and that of heavy object be $m _h$ such that $m _h>m _g$
Now during the equilibrium position we have the moments about the fulcrum as
$m _h\times \displaystyle\frac{l}{2}+y\times \displaystyle\frac{l}{4}=m _g\times \displaystyle\frac{l}{2}+x\times \displaystyle\frac{l}{4}$
or
$m _h-m _g=\displaystyle\frac{x-y}{2}$
Now when both the girl and the heavy object move half the distance we see that the moment about the pivot by the girl and the heavy object are $m _g\times \displaystyle\frac{l}{4}+x\times \displaystyle\frac{l}{4}$ and $m _h\times \displaystyle\frac{l}{4}+y\times \displaystyle\frac{l}{4}$

Subtracting moment of the girl from the moment of the heavy object
$(m _h+y)\displaystyle\frac{l}{4}-(m _g+x)\displaystyle\frac{l}{4}$
or
$\displaystyle\frac{l}{4}(m _h-m _g+y-x)$
or
$\displaystyle\frac{l}{4}(\displaystyle\frac{x}{2}-\displaystyle\frac{y}{2}+y-x)$
or
$\displaystyle\frac{l}{4}(\displaystyle\frac{y-x}{2})$
We get this value as negative because $x>y$
Thus the moment applied by the girl is more than that of the heavy object implying that the side of the girl tilts downwards

Three weight $W, 2W$ and $3W$, are connected to identical springs suspended from rigid horizontal rod. The assembly of the rod and the weights fall freely. The positions of the weights from the rod are such that

physics rotational motion of a rigid body and moment of inertia motion of rigid body rigid body equilibrium of a rigid body

  1. $3W$ will be farthest

  2. $W$ will be farthest

  3. all will be at the same distance

  4. $2W$ will be farthest

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Correct Option: C
Explanation:

Since all the weights fall freely under gravity along with the assembly of the rod, the acceleration of all objects is equal to $g$. Hence all the weights have same relative distance between them at all points, and hence no stretching of spring occurs and thus the position(distance of falling) of all the weights would be the same.

Assertion (A) : A wheel may be rotated with uniform angular velocity even though the tangential forces are applied on it.
Reason (R) : Angular acceleration of wheel is zero when tangential force and frictional force produce torques equal in magnitude and opposite in direction.

physics rotational motion of a rigid body and moment of inertia motion of rigid body rigid body equilibrium of a rigid body

  1. Both A and R are true and R is correct explanation of A

  2. Both A and R are true and R is not correct explanation of A

  3. A is true and R is false

  4. A is false and R is true

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Correct Option: A
Explanation:

A wheel is rotated with uniform angular velocity even if the tangential force is applied it happens only if a counter torque is acted on the wheel to balance the torque of the applied force. The $ angular$ $ acceleration $ of the wheel is $ zero$ if the torque of the tangential force is balanced by any other force $(i.e. friction)$.

A uniform rule is pivoted at its mid point. A weight of $50\ gf$ is suspended at one end of it. Where should a weight of $100\ gf$ be suspended, to keep the rule horizontal?

physics rotational motion of a rigid body and moment of inertia motion of rigid body rigid body equilibrium of a rigid body

  1. $10\ cm$ from mid point

  2. $25\ cm$ from mid point

  3. $20\ cm$ from mid point

  4. $15\ cm$ from mid point

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Correct Option: B
Explanation:

To keep the rod horizontal, total torque should be zero

$\Rightarrow \tau = 50gf (0.5) - 100gf (x) = 0$
$\Rightarrow 50gf (0.5) = 100gf (x) $
$\Rightarrow x= 0.25 $
So weight should suspended from $25 cm$ from mid point.
Therefore, B is correct option.

State whether true or false.
A cone resting on its side is an example of a body in neutral equilibrium.

physics rotational motion of a rigid body and moment of inertia motion of rigid body rigid body equilibrium of a rigid body

  1. True

  2. False

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Correct Option: A
Explanation:

Since, The energy of cone resting on its side is minimum, it will be in neutral equilibrium.

Therefore,statement is true.

State whether true or false.
Self balancing toys have curved and heavy bases.

physics rotational motion of a rigid body and moment of inertia motion of rigid body rigid body equilibrium of a rigid body

  1. True

  2. False

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Correct Option: A
Explanation:

Since curved and heavy base provide stability to the toys so they have curved and heavy bases.

Therefore, Given statement is true.

Calculate the force required to lift a load of $60\ N$, placed at a distance of $3\ m$, from the fulcrum if the effort force is applied at a distance of $6\ cm$ from the fulcrum.

physics rotational motion of a rigid body and moment of inertia motion of rigid body rigid body equilibrium of a rigid body

  1. $300\ N$

  2. $3000\ N$

  3. $1500\ N$

  4. $30\ N$

Show answer
Correct Option: B
Explanation:

In order to lift the load we have to apply force such that it can balance the torque by load.

Or, Total torque shold be zero.
We know  $\tau = r \times F$
$\Rightarrow \tau _{total} = (3)60 - (0.06)F$... where F is force required.
$\Rightarrow (3)60 = (0.06)F$

$\Rightarrow F = 3000N$
Therefore, B is correct option.

A body is in pure rotation. The linear speed $v$ of the particle, the distance $r$ of the particle from the axis and the angular velocity $\omega$ of the body are related as $\omega=\dfrac{v}{r}$. Thus

physics rotational motion of a rigid body and moment of inertia motion of rigid body rigid body equilibrium of a rigid body

  1. $\omega \propto \dfrac{1}{r}$

  2. $\omega \propto r$

  3. $v$

  4. $\omega$ $is\ independent\ of$ $r$

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Correct Option: A