Tag: equilibrium of a rigid body

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Questions Related to equilibrium of a rigid body

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

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

  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 $h$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a sphere on a smooth surface, the acceleration of the center of mass is given by a = F/m, according to Newton's second law. This value is independent of the point of application 'h'.

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

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?

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

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

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

  1. $3W$ will be farthest

  2. $W$ will be farthest

  3. all will be at the same distance

  4. $2W$ will be farthest
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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.

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(i.e. friction)$.

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

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?

  1. $10\ cm$ from mid point

  2. $25\ cm$ from mid point

  3. $20\ cm$ from mid point

  4. $15\ cm$ from mid point
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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.

  1. $300\ N$

  2. $3000\ N$

  3. $1500\ N$

  4. $30\ N$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

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

  2. $\omega \propto r$

  3. $v$

  4. $\omega$ $is\ independent\ of$ $r$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relationship v = omega * r implies that for a constant linear speed v, omega is inversely proportional to r (omega = v/r).