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Step-by-Step Solution
Step 1: Understand the Problem
A circular disc of mass M and radius R is initially rotating with angular speed \omega_1 . Another disc with the same mass M but half the radius ( \frac{R}{2} ) is placed coaxially on top of the first disc, initially at rest. After some time, both discs rotate together with a common angular speed \omega_2 . We need to find the percentage loss in the rotational kinetic energy during this process.
Step 2: Identify the Relevant Physical Principles
• Conservation of angular momentum applies because no external torque acts on the combined system during the slipping process.
• Kinetic energy may not remain constant because frictional forces act between the discs, causing energy dissipation.
Step 3: Express the Initial and Final Moments of Inertia
Let I_i be the moment of inertia of the initial rotating disc:
I_i = \frac{1}{2} M R^2 \,.
When the second disc is placed on top, the final moment of inertia I_f is the sum of both discs’ moments of inertia (since they rotate together about the same axis):
I_f = \frac{1}{2} M R^2 + \frac{1}{2} M \left(\frac{R}{2}\right)^2
= \frac{1}{2} M R^2 + \frac{1}{2} M \frac{R^2}{4}
= \frac{1}{2} M R^2 + \frac{1}{8} M R^2
= \frac{4}{8} M R^2 + \frac{1}{8} M R^2
= \frac{5}{8} M R^2
= \frac{5}{4} \left(\frac{1}{2} M R^2\right).
Step 4: Apply Conservation of Angular Momentum
Before the second disc is dropped, the total angular momentum is:
L_i = I_i \omega_1.
After they rotate together, the total angular momentum is:
L_f = I_f \omega_2.
Because angular momentum is conserved ( L_i = L_f ), we have:
I_i \omega_1 = I_f \omega_2
\quad \Longrightarrow \quad
\frac{1}{2} M R^2 \, \omega_1 = \frac{5}{8} M R^2 \, \omega_2.
Cancelling out the common factors, we get:
\omega_2 = \frac{4}{5} \, \omega_1.
Step 5: Calculate the Initial and Final Kinetic Energies
Initial Kinetic Energy:
\displaystyle K_i = \frac{1}{2} I_i \,\omega_1^2
= \frac{1}{2} \left(\frac{1}{2} M R^2\right) \omega_1^2
= \frac{1}{4} M R^2 \,\omega_1^2.
Final Kinetic Energy:
\displaystyle K_f = \frac{1}{2} I_f \,\omega_2^2
= \frac{1}{2} \left(\frac{5}{8} M R^2\right) \left(\frac{4}{5} \omega_1\right)^2.
= \frac{5}{16} M R^2 \times \frac{16}{25}\,\omega_1^2
= \left(\frac{5}{16} \times \frac{16}{25}\right) M R^2 \,\omega_1^2
= \frac{5 \times 16}{16 \times 25} M R^2 \,\omega_1^2
= \frac{5 \times 16}{400} M R^2 \,\omega_1^2
= \frac{80}{400} M R^2 \,\omega_1^2
= \frac{1}{5} M R^2 \,\omega_1^2.
Step 6: Determine the Percentage Loss in Kinetic Energy
The loss in kinetic energy is K_i - K_f , and the percentage loss is given by:
\%\text{loss}
= \frac{K_i - K_f}{K_i} \times 100\%.
Substitute K_i = \tfrac{1}{4} M R^2 \omega_1^2 and K_f = \tfrac{1}{5} M R^2 \omega_1^2 :
\%\text{loss}
= \frac{\tfrac{1}{4} M R^2 \omega_1^2 - \tfrac{1}{5} M R^2 \omega_1^2}{\tfrac{1}{4} M R^2 \omega_1^2} \times 100\%
= \left(1 - \frac{4}{5}\right) \times 100\%
= \frac{1}{5} \times 100\%
= 20\%.
Answer: 20%
