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A uniform circular disc A of radius r is...

A uniform circular disc A of radius r is made from A copper plate of thickness t and another uniform circular disc B of radius 2r is made from a copper plate of thickness t/2. The relation between the moments of inertia `I_A and I_B` is?

A

`I_AgtI_B`

B

`I_A=I_B`

C

`I_AltI_B`

D

Depends on the values of t and r.

Text Solution

AI Generated Solution

The correct Answer is:
To find the relationship between the moments of inertia \( I_A \) and \( I_B \) for the two discs, we will follow these steps: ### Step 1: Understand the Moment of Inertia Formula The moment of inertia \( I \) for a uniform circular disc is given by the formula: \[ I = \frac{1}{2} m r^2 \] where \( m \) is the mass of the disc and \( r \) is its radius. ### Step 2: Calculate the Mass of Disc A For disc A: - Radius \( r_A = r \) - Thickness \( t_A = t \) The volume \( V_A \) of disc A is given by: \[ V_A = \pi r_A^2 t_A = \pi r^2 t \] The mass \( m_A \) of disc A can be calculated using the density \( \rho \): \[ m_A = \rho V_A = \rho (\pi r^2 t) = \pi \rho r^2 t \] ### Step 3: Calculate the Moment of Inertia for Disc A Using the mass calculated above: \[ I_A = \frac{1}{2} m_A r_A^2 = \frac{1}{2} (\pi \rho r^2 t) r^2 = \frac{1}{2} \pi \rho t r^4 \] ### Step 4: Calculate the Mass of Disc B For disc B: - Radius \( r_B = 2r \) - Thickness \( t_B = \frac{t}{2} \) The volume \( V_B \) of disc B is given by: \[ V_B = \pi r_B^2 t_B = \pi (2r)^2 \left(\frac{t}{2}\right) = \pi (4r^2) \left(\frac{t}{2}\right) = 2 \pi r^2 t \] The mass \( m_B \) of disc B can be calculated using the density \( \rho \): \[ m_B = \rho V_B = \rho (2 \pi r^2 t) = 2 \pi \rho r^2 t \] ### Step 5: Calculate the Moment of Inertia for Disc B Using the mass calculated above: \[ I_B = \frac{1}{2} m_B r_B^2 = \frac{1}{2} (2 \pi \rho r^2 t) (2r)^2 = \frac{1}{2} (2 \pi \rho r^2 t) (4r^2) = 4 \pi \rho t r^4 \] ### Step 6: Find the Ratio of Moments of Inertia Now, we can find the ratio \( \frac{I_B}{I_A} \): \[ \frac{I_B}{I_A} = \frac{4 \pi \rho t r^4}{\frac{1}{2} \pi \rho t r^4} = \frac{4}{\frac{1}{2}} = 8 \] ### Conclusion Thus, the relationship between the moments of inertia is: \[ I_B = 8 I_A \]
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