A wheel has \(I=0.50\,\mathrm{kg\,m^2}\) and net torque \(3.0\,\mathrm{N\,m}\). Find \(\alpha\).
Question 2
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A tangential \(20\,\mathrm{N}\) force acts at radius \(0.15\,\mathrm{m}\). Find the torque magnitude.
Question 3
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If positive torque is counterclockwise and \(\alpha=-4\,\mathrm{rad\,s^{-2}}\), what is the sense of angular acceleration?
Question 4
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The same net torque is applied to two wheels. Wheel A has twice the moment of inertia of wheel B. Compare their angular accelerations.
Question 5
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A disk starts from rest and has constant angular acceleration \(5.0\,\mathrm{rad\,s^{-2}}\) for \(3.0\,\mathrm{s}\). Find its final angular velocity.
Question 6
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A pulley has \(I=0.20\,\mathrm{kg\,m^2}\) and radius \(0.10\,\mathrm{m}\). A string tension of \(8.0\,\mathrm{N}\) acts tangentially. Find \(\alpha\).
Question 7
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Two torques act on a wheel: \(6.0\,\mathrm{N\,m}\) counterclockwise and \(2.5\,\mathrm{N\,m}\) clockwise. If \(I=0.70\,\mathrm{kg\,m^2}\), find \(\alpha\).
Question 8
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A no-slip string unwinds from a pulley of radius \(0.12\,\mathrm{m}\). If the string acceleration is \(1.8\,\mathrm{m\,s^{-2}}\), find \(\alpha\).
Question 9
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A \(2.0\,\mathrm{kg}\) mass hangs from a light string wrapped around a fixed pulley with \(I=0.050\,\mathrm{kg\,m^2}\) and \(R=0.10\,\mathrm{m}\). If the mass accelerates downward at \(1.6\,\mathrm{m\,s^{-2}}\), find the string tension and the pulley's angular acceleration.
Question 10
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A wheel with \(I=1.2\,\mathrm{kg\,m^2}\) has angular speed \(10\,\mathrm{rad\,s^{-1}}\). A constant opposing torque of \(3.0\,\mathrm{N\,m}\) acts until it stops. Find the angular deceleration and stopping time.
Question 11
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A mass \(m\) hangs from a light string wound around a fixed pulley of radius \(R\) and moment of inertia \(I\). The string does not slip and the system is released from rest. Derive the mass's downward acceleration and the string tension.
Question 12
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Two hanging masses \(m_1\) and \(m_2\) are attached to light strings wound on different radii \(r_1\) and \(r_2\) of the same fixed pulley with moment of inertia \(I\). Taking positive \(\alpha\) as \(m_2\) moving downward, derive \(\alpha\), the two tensions, and the condition for the assumed direction to be correct.