Online Test: System of Particles and Rotational Motion | Class 11 Physics

Question 1 of 18

The centre of mass of a system is the:

heaviest particle

mass-weighted average position

geometric corner

fastest particle

Explanation: By definition $\vec{r}_{cm}=\frac{\sum m_i \vec{r}_i}{\sum m_i}$.

Question 2 of 18

Masses 1 kg and 4 kg are 10 m apart. The CM is at a distance from the 1 kg mass of:

$2\ \text{m}$

$5\ \text{m}$

$8\ \text{m}$

$10\ \text{m}$

Explanation: $x_{cm}=\frac{4\times10}{5}=8\ \text{m}$ from the 1 kg mass.

Question 3 of 18

Total momentum of a system of particles equals:

$M\vec{a}_{cm}$

$M\vec{v}_{cm}$

$\sum m_i$

$M\vec{r}_{cm}$

Explanation: $\vec{P}=M\vec{v}_{cm}$.

Question 4 of 18

The centre of mass moves under the influence of:

internal forces

net external force only

no force

friction only

Explanation: $M\vec{a}_{cm}=\vec{F}_{ext}$; internal forces cancel.

Question 5 of 18

The magnitude of torque is given by:

$rF$

$rF\cos\theta$

$rF\sin\theta$

$rF\tan\theta$

Explanation: $\vec{\tau}=\vec{r}\times\vec{F}$, so $\tau=rF\sin\theta$.

Question 6 of 18

The SI unit of torque is:

$\text{N}$

$\text{J}$

$\text{N m}$

$\text{W}$

Explanation: Torque is measured in newton metre.

Question 7 of 18

A force of 10 N acts perpendicular at 0.5 m from the axis. The torque is:

$2\ \text{N m}$

$5\ \text{N m}$

$10\ \text{N m}$

$20\ \text{N m}$

Explanation: $\tau=rF\sin90^\circ=0.5\times10=5\ \text{N m}$.

Question 8 of 18

Moment of inertia is given by:

$\sum m_i r_i$

$\sum m_i r_i^2$

$\sum m_i^2 r_i$

$\sum m_i/r_i$

Explanation: $I=\sum m_i r_i^2$.

Question 9 of 18

The moment of inertia of a solid disc of mass $M$, radius $R$ about its axis is:

$MR^2$

$\frac{1}{2}MR^2$

$\frac{2}{5}MR^2$

$\frac{1}{12}MR^2$

Explanation: For a solid disc/cylinder $I=\frac{1}{2}MR^2$.

Question 10 of 18

The radius of gyration $k$ is defined by:

$I=Mk$

$I=Mk^2$

$I=M/k^2$

$I=k/M$

Explanation: $I=Mk^2$, so $k=\sqrt{I/M}$.

Question 11 of 18

Angular momentum of a rigid body about a fixed axis is:

$I\alpha$

$I\omega$

$\frac{1}{2}I\omega^2$

$\tau\theta$

Explanation: $L=I\omega$.

Question 12 of 18

When the net external torque is zero, the conserved quantity is:

kinetic energy

angular momentum

torque

moment of inertia

Explanation: $\frac{dL}{dt}=\tau_{ext}=0$, so $L$ is constant.

Question 13 of 18

An ice-skater pulling in the arms while spinning will:

slow down

stop

spin faster

stay the same

Explanation: $I$ decreases, so $\omega$ increases to keep $L=I\omega$ constant.

Question 14 of 18

The rotational analogue of Newton's second law is:

$L=I\omega$

$\tau=I\alpha$

$v=\omega R$

$KE=\frac{1}{2}I\omega^2$

Explanation: $\tau=I\alpha$ mirrors $F=ma$.

Question 15 of 18

The rotational kinetic energy of a body is:

$I\omega$

$\frac{1}{2}I\omega^2$

$\frac{1}{2}I\omega$

$I\omega^2$

Explanation: $KE_{rot}=\frac{1}{2}I\omega^2$.

Question 16 of 18

For rolling without slipping, the centre velocity satisfies:

$v=\omega/R$

$v=\omega R$

$v=R/\omega$

$v=\omega R^2$

Explanation: The contact point is at rest, giving $v=\omega R$.

Question 17 of 18

The total kinetic energy of a rolling body is:

$\frac{1}{2}mv^2$

$\frac{1}{2}I\omega^2$

$\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$mv^2+I\omega^2$

Explanation: Rolling KE is the sum of translational and rotational parts.

Question 18 of 18

A solid sphere, a disc and a ring roll down the same incline. The one reaching the bottom first is the:

ring

disc

solid sphere

all together

Explanation: The sphere has the smallest $k^2/R^2$, hence the largest acceleration.

Online Test — System of Particles and Rotational Motion