Answer :
Final Answer:
Answer 1: The mass [tex]\(M\)[/tex] of the meter stick is [tex]\(6 \, \text{kg}\)[/tex].
Answer 2: To achieve equilibrium when [tex]\(m_1 > m_2\)[/tex], the rod should be supported at a point [tex]\(x\)[/tex] where [tex]\(x = \frac{m_2}{m_1 + m_2} \times L\)[/tex].
Answer 3: The linear speeds of the solid disk and hoop at the bottom of the incline are equal due to the conservation of energy.
Answer 4: The kinetic energy of rotation of the rotating rod is [tex]\( \frac{1}{3} M L^2 \omega^2 \).[/tex]
Answer 5: The velocity of the center of mass of the sphere when it reaches the bottom of the incline is [tex]\( \sqrt{\frac{2gh}{1 + \frac{2}{5}}}\)[/tex].
Explanation:
Answer 1: To find [tex]\(M\)[/tex], apply the principle of torque equilibrium: [tex]\(\sum \tau = 0\)[/tex]. The torques due to the masses must balance. The torque from the 2 kg mass [tex](\(20 \, \text{cm}\))[/tex] is [tex]\(2 \times 9.8 \times 0.2 \, \text{m} = 3.92 \[/tex], [tex]\text{Nm}\)[/tex], and from the 3 kg mass [tex](\(90 \, \text{cm}\))[/tex] is [tex]\(3 \times 9.8 \times 0.3 \, \text{m} = 8.82 \, \text{Nm}\)[/tex]. For equilibrium, [tex]\(3.92 + M \times 9.8 \times 0.6 = 8.82\), yielding \(M = 6 \, \text{kg}\)[/tex].
Answer 2: For equilibrium, the torques due to [tex]\(m_1\) and \(m_2\)[/tex] must balance. Using [tex]\(\sum \tau = 0\), \(m_1 \times x = m_2 \times (L - x)\)[/tex], solving for [tex]\(x\)[/tex] gives [tex]\(x = \frac{m_2}{m_1 + m_2} \times L\).\\[/tex]
Answer 3: Applying conservation of energy, the potential energy at the top is converted into kinetic energy at the bottom. Equating the energies of the hoop and disk, the linear speeds are the same.
Answer 4: The kinetic energy of rotation for a thin rod rotating about its end is [tex]\( \frac{1}{3} M L^2 \omega^2 \).[/tex]
Answer 5: Using energy conservation, the potential energy at the top converts to translational and rotational kinetic energy at the bottom, yielding the velocity formula [tex]\( \sqrt{\frac{2gh}{1 + \frac{2}{5}}}\)[/tex].
