A mass 9.3 kg is attached to a light cord, which is wound around a pulley. The p

Question

A mass 9.3 kg is attached to a light cord, which is wound around a pulley. The pulley is a uniform solid cylinder of radius 6.7 cm and mass 3 kg. The acceleration of gravity is 9.8 m/s^2. (1.) What is the magnitude of the net torque on the system about the point on the axis at the center of the pulley? Answer in units of N m. When the mass has a speed v = 27 m/s, the pulley has an angular speed w = v/R. (2.) Determine the magnitude of the total angular momentum of the system about a point on the axis at the center of the pulley? Answer in units of kg m^2/s. (3.) Using fact that  = dL/dt and your result from the previous section, calculate the magnitude of the acceleration of the mass. Answer in units of m/s^2.

Explanation / Answer

1. The equation of the system is mg - /r = ma where torque = I tehn =/I =a/r and I=0.5m2r2 a= r solving for torque m1g - /r = m1 r m1g - /r = m1 (/I)r (1/r + m1r/I) =m1g since I=0.5m2r2  
= m1g/ ( 1/r + m1 r /[0.5m2 r2 ]) = m1g r/ ( 1 + 2m1  /m2 ) = 9.3 x 9.8 x 0.067/ ( 1 + 2x9.3  /3.0) = 0.85 Nm - 2) L= I = V/r I= 0.5m2r2  we have
L= V 0.5m2r2  /r L= 0.5 Vm2r L= 0.5 x 27 x 3.0 x 0.067 L=2.71 kg m2/s - - 3) Since =dL/dt and = I dL/dt =dL/dt =/I = 0.85 /[0.5 x 3.0 x0.0672] = 126 rad/s

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