Problem 43 - Master Resource Book in JEE Main Physics

Question

Let \(M\) and \(R\) be the mass of radius of the solid sphere and hollow cylinder. Moment of inertia of the hollow cylinder about its axis of symmetry, $$ I_{1}=M R^{2} $$ Moment of linear of the solid sphere about its diameter $$ I_{2}=\frac{2}{5} M R^{2} $$ Let torque \(\tau\) of triangle magnitude be applied on hollow cylinder and solid sphere. the angular acceleration produced in it are \(\alpha_{1}\) and \(\alpha_{2}\) respectively. \(\therefore\) \(\tau=I_{1} \alpha_{1}\) and \(\quad \tau=I_{2} \alpha_{2}\) Therefore, $$ \begin{gathered} I_{1} \alpha_{1}=I_{2} \alpha_{2} \\ \frac{\alpha_{1}}{\alpha_{2}}=\frac{l_{2}}{I_{1}}=\frac{\frac{2}{5} M R^{2}}{M R^{2}}=\frac{2}{5} \end{gathered} $$ or or $$ \alpha_{2}=\frac{5}{2} \alpha_{1} $$ $$ =2.5 \alpha_{1} $$ Let after time \(t, \omega_{1}\) and \(\omega_{2}\) be the angular speeds of the hollow cylinder and solid sphere respectively. \(\therefore\) $$ \omega_{1}=\omega_{0}+\alpha_{1} t $$ and $$ \begin{aligned} &\omega_{2}=\omega_{0}+\alpha_{2} t \\ &=\omega_{0}+2.5 \alpha_{1} t \end{aligned} $$ From Eqs. (ii) and (iii), we get $$ \omega_{2}>\omega_{1} $$ Therefore, solid sphere will acquire a greater angular speed after a given time.

Step-by-Step Solution

Answer

The solid sphere will have a greater angular speed than the hollow cylinder.

Step 1: Identify Given Variables and Known Equations

We are given the following properties for the hollow cylinder and solid sphere:- Moment of inertia of hollow cylinder: \( I_1 = M R^2 \)- Moment of inertia of solid sphere: \( I_2 = \frac{2}{5} M R^2 \).We are also given the torques \( \tau\) applied, which result in angular accelerations \( \alpha_1 \) for the cylinder and \( \alpha_2 \) for the sphere:- \( \tau = I_1 \alpha_1 \)- \( \tau = I_2 \alpha_2 \).

Step 2: Relate Angular Accelerations

Use torque equations to set up the relationship between \( \alpha_1 \) and \( \alpha_2 \):\[ I_1 \alpha_1 = I_2 \alpha_2 \]Substituting expressions for \(I_1\) and \(I_2\):\[ M R^2 \alpha_1 = \frac{2}{5} M R^2 \alpha_2 \].Cancel \( M R^2 \) from both sides:\[ \alpha_1 = \frac{2}{5} \alpha_2 \].Solve for \( \alpha_2 \):\[ \alpha_2 = \frac{5}{2} \alpha_1 \].

Step 3: Calculate Angular Speeds

Given the initial angular speed \( \omega_0 \) and the relation \( \alpha_2 = 2.5 \alpha_1 \), calculate the angular speeds after time \( t \).- For hollow cylinder: \( \omega_1 = \omega_0 + \alpha_1 t \).- For solid sphere: \( \omega_2 = \omega_0 + 2.5 \alpha_1 t \).

Step 4: Compare Angular Speeds

Comparing \( \omega_2 \) and \( \omega_1 \) after time \( t \):- \( \omega_1 = \omega_0 + \alpha_1 t \)- \( \omega_2 = \omega_0 + 2.5 \alpha_1 t \)Since \( 2.5 \alpha_1 t > \alpha_1 t \), we have \( \omega_2 > \omega_1 \).

Step 5: Conclusion

Therefore, the solid sphere will acquire a greater angular speed than the hollow cylinder after the same amount of time under the same torque.

Key Concepts

Moment of Inertia

Moment of inertia, often termed the "rotational inertia," plays a crucial role in understanding the resistance an object offers against its rotation. It depends on both the mass distribution of the object and the axis about which it rotates.
For the hollow cylinder, its moment of inertia is defined as \( I_1 = MR^2 \). This formula indicates that the mass \( M \) is concentrated at a distance \( R \) from the axis, consequently providing a higher inertia compared to objects with a similar mass positioned closer to the axis.
Meanwhile, the solid sphere's moment of inertia is \( I_2 = \frac{2}{5} MR^2 \). Here, the \( \frac{2}{5} \) factor suggests that the mass of the solid sphere is distributed closer to the axis—giving it a lower moment of inertia than the hollow cylinder.
Understanding the moment of inertia helps to predict how these objects react when torques of the same magnitude are applied. It's one of the foundational concepts you need to grasp when dealing with problems involving rotation.

Torque

Torque is a physical concept similar to force but for rotational motion. It effectively determines how much a force acting on an object causes the object to rotate. The effectiveness of this force is heightened when it acts further from the axis of rotation. The formula for torque is expressed as \( \tau = I \alpha \), where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.
In our exercise, the same torque \( \tau \) is applied to both the hollow cylinder and the solid sphere. Despite \( \tau \) being the same, the angular accelerations \( \alpha_1 \) and \( \alpha_2 \) differ due to the differing moments of inertia. This phenomenon underscore how crucial moment of inertia is for determining an object's response to applied torque.
Thus, an understanding of torque not only opens the door to comprehending forces in linear motion transformed to rotational scenarios but also to manipulating rotational effects like acceleration and speed.

Angular Speed

Angular speed refers to how fast an object rotates or revolves relative to another point, essentially evaluating changes in position angle over time. It is significant when comparing rotational behaviors of objects subjected to identical conditions of force and torque.
In the context of our problem, after a certain time \( t \), the angular speeds of both the cylinder, \( \omega_1 \), and the sphere, \( \omega_2 \), are calculated using initial angular speeds \( \omega_0 \) and the respective angular accelerations. For the cylinder, \( \omega_1 = \omega_0 + \alpha_1 t \), while for the sphere it follows \( \omega_2 = \omega_0 + 2.5 \alpha_1 t \).
From the equations, \( \omega_2 \) is greater than \( \omega_1 \), resulting in the sphere reaching a higher angular speed due to its greater angular acceleration. The comparison of their angular speeds further highlights how the different properties of the objects determine dynamic outcomes in such scenarios of rotational motion.