Question

An electric motor turns a flywheel through a drive belt that joins a pulley on the motor and a pulley that is rigidly attached to the flywheel as shown in figure. The flywheel is a solid disc with a mass of $80\cdot 0Kg$ and a radius $R=0\cdot 50m$. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of $r=0\cdot 20m$ . The tension $T_u$ in the upper (taut) segment of the belt is $400N$, and the flywheel has a clockwise angular acceleration of $2\cdot 0rad/s^2$. The tension in the lower (slack) segment of the belt will be

• A. $75N$
• B. $90N$
• C. $300N$
• D. $180N$

Answer 1

The correct option is C $300N$

Step 1:Find the torque about the center of the fly wheel.
Given:
Mass of solid disc$\left(M\right)=80kg$
Radius of disc$\left(R\right)=0\cdot 50m$
Radius of pulley$\left(r\right)=0\cdot 20m$
Tension in the upper segment of belt$\left(T_u\right)=400N$
Angular acceleration$\left(\alpha \right)=2rad/s^2$
As per the diagram, flywheel is a solid disc of mass $M$ and radius $R$ with axis through its center.
As we know,Moment of inertia of the disc $I=\frac{\left(M{R}^2\right)}{2}$Hence, by applying torque equation about center of the flywheel we get,
$\Sigma \tau =I\alpha =T_{u}r-T_{b}r$
$\Sigma \tau =\frac{1}{2}M{R}^2\alpha$
Step 2:Find the tension in the lower segment $\left(T_b\right)$ of the fly wheel.
As we know,
$\Rightarrow T_b=T_u-\frac{\left(M{R}^2\alpha \right)}{2r}$
Therefore, by putting the values we get,
$T_b=400N-\frac{\left(80.0kg\right)(0.50m{)}^2\frac{2.0rad}{s^2}}{2\left(0.20m\right)}$
$T_b=300N$

Final Answer: $a$