Two identical masses are attached to frictionless pulleys by very light strings wrapped around the rim of the pulley and are released from rest. Both pulleys have the same mass and same diameter, but one is solid and the other is a hoop. As the masses fall, in which case is the tension in the string greater, or is it the same in both cases? Justify your answer

Tension is defined as the pulling force transferred axially by a rod, cable, chain, or other similar object, or by each end of a comparable three-dimensional object. It can also be thought of as the action-reaction pair of forces acting at each end of the aforementioned elements.

We have two identical masses that are attached to two frictionless pulleys by very light strings wrapped around the rim of the pulley.

The masses are released from rest. Both pulleys have the same mass and same diameter.

One of the pulleys is solid and the other is a hoop.

The free-body diagram shown in the figure below shows some of the forces acting on the block and then pulley;$m\overrightarrow{g}$is the gravitational force exerterd on the block acting downwards and $\overrightarrow{T}$is the tension in the string.

By, Newton’s second law, we know that

$\sum _{}F=ma$ (1)

Now, let downward be positive direction. Model the block as a particle under a net force in the vertical direction and apply equation (1). We get,

$$\begin{gathered} \sum _{}F=mg-T_1 \\ =ma \end{gathered}$$

$T_1=mg-m$ (2)

Now, Net torque is given by

$\sum {\tau }_{ext}=I\alpha$ (3)

Where, is moment of inertia and$\alpha$ is angular acceleration.

Also, magnitude of torque associated with force$\overrightarrow{F}$ acting on an object at a distance r from the rotation axis is;

$$\begin{gathered} \tau =rF\sin \theta \\ =Fd \end{gathered}$$ (4)

Where$\theta$ is angle between the position vector of the point of application of the force and force vector andd is the moment of arm of the force which is the perpendicular distance from the rotation axis to the line of action of the force.

Now, let clockwise be positive direction.

The torque on the disk about its center is due to the tension forceT only.

So, from equation (3)

$\sum _{}\tau =l\alpha =T_1$ (5)

Where, l is a moment of inertia of solid

Now, we know that moment of inertia of solid can be defined as,

$l=\frac{1}{2}M{R}^2$ (6)

WhereM is mass of pulley andR is radius of pulley.

Also When a rigid body rotates about a fixed axis, the angular acceleration are related to the transitional acceleration through the relationship. So we have,

$a_t=r\alpha$ (7)

Now, from equation (5), (6) and (7),

$$\begin{gathered} T_{1}R=\left(\frac{1}{2}M{R}^2\right)\left(\frac{a}{R}\right) \\ {T}_1=\frac{1}{2}Ma \end{gathered}$$

So we get,$a=\frac{2T_1}{M}$

Now, put value of in equation (2). We get,

$$\begin{gathered} T_1=mg-m\left(\frac{2T_1}{M}\right) \\ {T}_1\left(1+\frac{2}{M}\right)=mg \end{gathered}$$

$T_1=\frac{mg}{\left(1+{\displaystyle \frac{2}{M}}\right)}$ (8)

Similar to above step, apply equation (1), we get

$$\begin{gathered} \sum _{}{F}_y=mg-T_2 \\ =ma \end{gathered}$$

$T_2-mg-ma$ (9)

Model the hoop as a rigid object under a net torque and apply equation (3) and (4). We get.

$\sum \tau =I\alpha =T_2$ (10)

Where,l is a moment of inertia of hoop

Now, we know that moment of inertia of hoop can be defined as,

$I=M{R}^2$ (11)

Now, from equation (10), (11) and (6),

$$\begin{gathered} T_{2}R=\left(M{R}^2\right)\left(\frac{a}{R}\right) \\ {T}_2=Ma \end{gathered}$$

So we get,$a=\frac{T_2}{M}$

Now, put value of in equation (0). We get,

$$\begin{gathered} T_2=mg-m\left(\frac{T_2}{M}\right) \\ {T}_2\left(1+\frac{1}{M}\right)=mg \end{gathered}$$

$T_2=\frac{mg}{\left(1+{\displaystyle \frac{1}{M}}\right)}$ (12)

Now, from equation (8) and (12), we can see that $T_2>T_1$.

Hence, the hoop pulley has greater tension than the solid pulley