$A$ wheel is rotating about an axis through its centre at $720 \, rpm$. It is acted on by a constant torque opposing its motion for $8 \, s$ to bring it to rest finally. The value of torque (in $N \cdot m$) is (Given $I = \frac{24}{\pi} \, kg \cdot m^2$)

What constant force must be applied tangentially at the equator to stop the Earth's rotation in one day?

$A$ massless rod of length $2l$ has equal point masses $m$ attached to its two ends as shown in the figure. The rod is rotating about an axis passing through its centre and making an angle $\alpha$ with the axis. The magnitude of the rate of change of angular momentum of the rod,i.e.,$\left|\frac{dL}{dt}\right|$,is equal to:

$A$ uniform rod of mass $m$ and length $2l$ is balanced on a triangular prism. Now,a length of $l/2$ is cut from one end of the rod and placed over the shortened part such that the ends meet. The initial angular acceleration is:

$A$ mass of $10 \ kg$ is attached to one end of a massless rod and rotated in a circle of radius $30 \ cm$ with an angular velocity of $10 \ rad/s$. If the body is brought to rest in $10 \ s$ by applying a brake,the magnitude of the torque applied will be ...... $N-m$.

$A$ wheel with a moment of inertia of $5 \times 10^{-3} \ kg \cdot m^2$ is rotating at a rate of $20 \ rev/s$. The angular deceleration required to bring it to rest in $20 \ s$ is: