In the figure below, the incline is frictionless and the string passes through t
ID: 1911151 • Letter: I
Question
In the figure below, the incline is frictionless and the string passes through the center of mass of each block. The pulley has a moment of inertia I and radius R. (Use the following as necessary: v, ?, m1, m2, R, I, and g.) (a) Find the net torque acting on the system (the two masses, string, and pulley) about the center of the pulley. ?net = Rg(m2sin*-m1) Correct: Your answer is correct. . (b) Write an expression for the total angular momentum of the system about the center of the pulley. Assume the masses are moving with a speed v. L = Incorrect: Your answer is incorrect. . (c) Find the acceleration of the masses by using your results for Parts (a) and (b) and by setting the net torque equal to the rate of change of the system's angular momentum.Explanation / Answer
Given a figure with a incline with angle ? that has a block with mass m_2 sliding down down the incline with velocity v and the block is attatched to another block of mass m_1 hanging off the side of the incline by a string that passes through a pulley, the incline is frictionless and the string passes through the center of mass of each block. The pulley has a moment of inertia I and radius R. (Use the following variables as necessary: v, theta for ?, m_1 for m1, m_2 for m2, R, I, and g.) (a) Find the net torque acting on the system (the two masses, string, and pulley) about the center of the pulley (b) Write an expression for the total angular momentum of the system about the center of the pulley. Assume the masses are moving with a speed v. (c) Find the acceleration of the masses by using your results for Parts (a) and (b) and by setting the net torque equal to the rate of change of the system's angular momentum. A) The net torque is caused by the components of the force of gravity acting on both blocks. Define our coordinate system, such that the pulley spinning toward the sliding block is positive. The hanging block applies a torque of the following to the system: tau_hang = -m2*g*R The sliding block applies a torque (produced by the component of its weight which is along the incline) of the following to the system: tau_slide = m1*g*R*sin(theta) Net torque: tau_net = g*R*(m1*sin(theta) - m2) B) Angular momentum is contributed by the pulley, block 1 sliding, and block 2 rising. Contribution of pulley: L_pulley = I*omega Contribution of sliding block: L_slide = m1*v*R Contribution of rising free hanging block: L_slide = m2*v*R Total: Lnet = I*omega + (m1 + m2)*v*R No-slip condition, kinematics constraint: omega = v/R Thus our resulting expression is: Lnet = v*(I + (m1 + m2)*R^2)/R C) tau_net = d Lnet/ dt d Lnet/dt = dv/dt * (I + (m1 + m2)*R^2)/R, since only v varies with time dv/dt is by definition acceleration a, thus: tau_net = a*(I + (m1 + m2)*R^2)/R Recall expression for tau_net: g*R*(m1*sin(theta) - m2) = a*(I + (m1 + m2)*R^2)/R Solve for a, and we get a resulting expression: a = g*R^2*(m1*sin(theta) - m2)/(R^2*(m1 + m2) + I) referenece http://www.physicsforums.com/archive/index.php/t-254561.html