Answered

A solid, uniform disc of mass M and radius R is able to rotate

without friction in the vertical plane about a horizontal axis through

its center of mass. Two small but heavy point particles of mass m1

and m2, where m1 > m2, are fixed to the disc at the very edge on

each side. If the disc is allowed to freely rotate, what is the angular

acceleration α the moment it is released from this position? Be sure

to include the direction. For your diagram draw the relevant physical

quantities on the provided figure. Hint: How many different moments

of inertia do you need to consider?

Answer :

Answer:

Angular acceleration of the disc just after the release is given as

[tex]\alpha = \frac{(m_1g - m_2g)R}{\frac{1}{2}MR^2 + m_1R^2 + m_2R^2}[/tex]

Explanation:

Here we know that the disc is hinged about the center and hence the weight of two particles will apply torque on the disc in opposite directions

So here net torque on the disc at the given position is

[tex]\tau = (m_1g - m_2g)R[/tex]

now total moment of inertia of the system of the disc is given as

[tex]I = I_{disc} + I_{particle}[/tex]

[tex]I = \frac{1}{2}MR^2 + m_1 R^2 + m_2R^2[/tex]

now we have

[tex]\tau = I\alpha[/tex]

so we have

[tex](m_1g - m_2g)R = \frac{1}{2}MR^2 + m_1R^2 + m_2R^2)\alpha[/tex]

so we have

[tex]\alpha = \frac{(m_1g - m_2g)R}{\frac{1}{2}MR^2 + m_1R^2 + m_2R^2}[/tex]

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