Numerical on equilibrium of a rigid body, moments, and center of gravity for JEE and CBSE board exams

1. A uniform rod of length L and mass M is hinged at one end and can rotate freely in a vertical plane. The rod is released from a horizontal position. What is the magnitude of the torque acting on the rod just before it reaches the vertical position?

Solution:

$$At\space position \theta, $$ $$\tau_{net}=-mgsin\theta L/2$$ $$\alpha=-\dfrac{3g}{2L}sin\theta$$

$$\omega^2=\omega_0^2-\int_{0}^{\theta}2\alpha d\theta$$

$$\omega^2=\omega_0^2+\dfrac{3g}{L}cos\theta$$

At vertical position,$$\theta=\pi/2$$

Substituting the value of ω, we get, $$\omega^2=\omega_0^2-3g$$

At the initial condition $$\omega=0, \space thus \space \omega_0=\sqrt{3g}$$

So, the required torque is:

$$\tau=I\alpha=I\times \dfrac{d^2\theta}{dt^2}$$

$$=I\times\dfrac{d\omega}{dt}$$

$$=I\times(0-\omega_0)$$

$$=\frac{1}{3}ML^2\times (0-\sqrt{3g})$$

$$=-\frac{1}{3}\sqrt{3}M g L$$

2. A uniform disk of mass M and radius R is initially at rest on a horizontal surface. A constant force F is applied tangentially to the disk at a point on its rim. What is the magnitude of the angular acceleration of the disk?

Solution:

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

$$\alpha=\frac{\tau}{I}=\frac{FR}{\frac{1}{2}MR^2}=\frac{2F}{MR}$$

3. A sphere of mass M and radius R is in equilibrium on an inclined plane. The angle of the incline is θ. What is the magnitude of the normal force acting on the sphere?

Solution:

$$\Sigma F_x=ma_x$$ $$N\sin\theta-F=0$$ $$N=\dfrac{F}{\sin\theta}$$

$$\Sigma F_y=ma_y$$ $$N\cos\theta-Mg=0$$

Substituting the value of N from the first equation, we get:

$$F\cot\theta-Mg=0$$

$$F=Mg\cot\theta$$

4. A uniform cylinder of mass M and radius R is placed on a horizontal surface. A force F is applied to the cylinder at a point on its rim, making an angle θ with the horizontal. What is the magnitude of the acceleration of the center of mass of the cylinder?

Solution:

$$\Sigma F=ma$$ $$F\sin\theta=Ma$$ $$a=F\sin\theta/M$$

5. A pendulum consists of a bob of mass m attached to a string of length L. The pendulum is released from a horizontal position. What is the speed of the bob when it reaches the lowest point of its trajectory?

Solution:

$$Mechanical \space energy (at the highest point) =mgl$$ $$Kinetic\space energy + Potential \space energy (at the lowest point) =\dfrac{1}{2}mv^2+0$$

Using the conservation of energy principle, we get: $$mgl=\dfrac{1}{2}mv^2$$ $$v=\sqrt{2gl}$$