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A flywheel with moment of inertia of 2 $\phantom{\rule{2pt}{0ex}}\mathrm{kg}-{m}^{2}$ , is rotating at 10 rad/s about a perpendicular axis. The flywheel is brought to rest uniformly in 10 seconds due to friction at the axle. The magnitude of torque due to friction is (in N-m) :
$$\begin{array}{l}\mathrm{(a)}\phantom{\rule{4pt}{0ex}}0\phantom{\rule{4pt}{0ex}}\mathrm{(b)}\phantom{\rule{4pt}{0ex}}5\phantom{\rule{4pt}{0ex}}\mathrm{(c)}\phantom{\rule{4pt}{0ex}}10\phantom{\rule{4pt}{0ex}}\mathrm{(d)}\phantom{\rule{4pt}{0ex}}15\end{array}$$
The angular deceleration is uniform i.e. constant as given in the question. The torque due to friction at the axle is, therefore, also constant. Let “α” be the deceleration.
Applying, equation of motion for constant angular acceleration, we have :
$$\begin{array}{l}{\omega}_{f}={\omega}_{i}+\alpha t\end{array}$$
$$\begin{array}{l}\mathrm{Here,}\phantom{\rule{2pt}{0ex}}{\omega}_{i}=10\phantom{\rule{2pt}{0ex}}\mathrm{rad}/s;\phantom{\rule{2pt}{0ex}}{\omega}_{f}=0;t=2s,\end{array}$$
$$\begin{array}{l}\Rightarrow 0=10+\alpha \phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2\end{array}$$
$$\begin{array}{l}\Rightarrow \alpha =-5\phantom{\rule{2pt}{0ex}}rad/{s}^{2}\end{array}$$
The magnitude of torque is :
$$\begin{array}{l}\Rightarrow \tau =I\alpha =2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}5=10\phantom{\rule{2pt}{0ex}}N-m\end{array}$$
Hence, option (c) is correct.
A circular disk of mass 2 kg and radius 1 m is free to rotate about its central axis. The disk is initially at rest. Then, the constant tangential force (in Newton) required to rotate the disk with angular velocity 10 rad/s in 5 second is :
$$\begin{array}{l}\mathrm{(a)}\phantom{\rule{4pt}{0ex}}2\phantom{\rule{4pt}{0ex}}\mathrm{(b)}\phantom{\rule{4pt}{0ex}}4\phantom{\rule{4pt}{0ex}}\mathrm{(c)}\phantom{\rule{4pt}{0ex}}8\phantom{\rule{4pt}{0ex}}\mathrm{(d)}\phantom{\rule{4pt}{0ex}}16\end{array}$$
This question is similar to earlier one except that we need to find the tangential force instead of torque. Since torque is equal to the product of tangential force and moment arm, we need to find torque first and then find the required tangential force.
A constant tangential force results in a torque along the axis of rotation. The torque, in turn, produces a constant angular acceleration “α”. As such, we can apply equation of motion for constant angular acceleration,
$$\begin{array}{l}{\omega}_{f}={\omega}_{i}+\alpha t\end{array}$$
$$\begin{array}{l}\mathrm{Here,}{\omega}_{i}=0;\phantom{\rule{2pt}{0ex}}{\omega}_{f}=10\phantom{\rule{2pt}{0ex}}\mathrm{rad}/s;t=2s,\end{array}$$
$$\begin{array}{l}\Rightarrow 10=0+\alpha \phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2\end{array}$$
$$\begin{array}{l}\Rightarrow \alpha =-5\phantom{\rule{2pt}{0ex}}rad/{s}^{2}\end{array}$$
Applying Newton’s second law for rotation, we have :
$$\begin{array}{l}\tau =I\alpha \end{array}$$
Here,
$$\begin{array}{l}I=M{R}^{2}=2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{1}^{2}=2\phantom{\rule{2pt}{0ex}}\mathrm{Kg}-{m}^{2}\end{array}$$
Hence,
$$\begin{array}{l}\tau =I\alpha =2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2=4\phantom{\rule{2pt}{0ex}}N-m\end{array}$$
Now, torque is also given as :
$$\begin{array}{l}\tau =RF\end{array}$$
$$\begin{array}{l}F=\frac{\tau}{R}=\frac{4}{1}=4\phantom{\rule{2pt}{0ex}}N\end{array}$$
Hence, option (b) is correct.
A wheel of 10 kg, mounted on its central axis, is acted upon by four forces. The lines of action of forces are in the plane perpendicular to the axis of rotation as shown in the figure. If the magnitudes of forces and their directions during rotation are same, then (considering, R = 10 m) :
$$\begin{array}{l}\mathrm{(a)}\phantom{\rule{4pt}{0ex}}\text{the magnitude of net torque is zero}\\ \phantom{\rule{4pt}{0ex}}\mathrm{(b)}\phantom{\rule{4pt}{0ex}}\text{the magnitude of net acceleration is 2}\mathrm{rad}/{s}^{2}\\ \phantom{\rule{4pt}{0ex}}\mathrm{(c)}\phantom{\rule{4pt}{0ex}}\text{the magnitude of net acceleration is 4}\mathrm{rad}/{s}^{2}\\ \phantom{\rule{4pt}{0ex}}\mathrm{(d)}\phantom{\rule{4pt}{0ex}}\text{the direction of rotation is anti-clockwise}\end{array}$$
The forces of 3N and 5N act through the axis of rotation. They do not constitute torque. The force of 4N constitute a clockwise torque of 4 x 10 = 40 N-m. On the other hand, the force of 2N constitute an anticlockwise torque of 2 x 20 = 40 N-m.
The cylinder, therefore, is acted by two equal, but opposite torques. Therefore, the net torque on the body is zero.
Hence, option (a) is correct.
Two flywheels of moments of inertia 1 and 2 $\phantom{\rule{2pt}{0ex}}\mathrm{kg}-{m}^{2}$ and radii 1 and 2 meters respectively are connected by a belt as shown in the figure. The angular speed of larger wheel is increased at a constant rate 2 $\phantom{\rule{2pt}{0ex}}\mathrm{rad}/{s}^{2}$ without any slippage between wheels and the belt. Then,
$$\begin{array}{l}\mathrm{(a)}\text{Tension in the belt is same}\\ \phantom{\rule{4pt}{0ex}}\mathrm{(b)}\phantom{\rule{4pt}{0ex}}\text{Linear velocity of the belt is same everywhere}\\ \phantom{\rule{4pt}{0ex}}\mathrm{(c)}\phantom{\rule{4pt}{0ex}}\text{Torques on the wheels are same}\\ \phantom{\rule{4pt}{0ex}}\mathrm{(d)}\phantom{\rule{4pt}{0ex}}\text{Angular accelerations of the wheels are same}\end{array}$$
Since, no slippage between wheels and the belt is involved, the linear velocity of the belt is same through out.
The rotation of the larger wheel causes the smaller wheel to rotate as well. Since two wheels are accelerated from at constant rate, it means that they are acted upon by constant torques. On the other hand, a net tangential force on either of the wheels is required to constitute a torque. As such, tension in the belt needs to be different so that there is a net tangential force on the flywheel.
Let ω be the angular speed of the larger wheel. The corresponding linear speed of the belt is :
$$\begin{array}{l}v={\omega}_{1}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{r}_{1}=\omega \phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2=2\omega \end{array}$$
The angular speed of the smaller wheel is :
$$\begin{array}{l}{\omega}_{1}=\frac{v}{{r}_{2}}=\frac{2\omega}{1}=2\omega \end{array}$$
On the other hand, the linear acceleration of the belt is :
$$\begin{array}{l}a={\alpha}_{1}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{r}_{1}=2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2=4\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$$
The angular acceleration of the smaller wheel is :
$$\begin{array}{l}\Rightarrow {\alpha}_{2}=\frac{a}{{r}_{2}}=\frac{4}{1}=4\phantom{\rule{2pt}{0ex}}\mathrm{rad}/{s}^{2}\end{array}$$
Thus, angular accelerations of the wheels are not same. Now, the torque on the larger wheel is :
$$\begin{array}{l}\Rightarrow {\tau}_{1}={I}_{1}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{\alpha}_{1}=2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2=4\phantom{\rule{2pt}{0ex}}N-m\end{array}$$
The torque on the smaller wheel is :
$$\begin{array}{l}\Rightarrow {\tau}_{2}={I}_{2}\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{\alpha}_{2}=1\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}4=4\phantom{\rule{2pt}{0ex}}N-m\end{array}$$
Thus, the torques on the wheels is same.
Hence, options (b) and (c) are correct.
1. (a) 2. (a), (b) and (d) 3. (c) 4. (b) 5. (a)
6. (b) and (c)
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