(b) Suppose the pulley has mass m p and radius R . Find the acceleration of m 1
ID: 2244524 • Letter: #
Question
(b) Suppose the pulley has mass mp and radius R. Find the acceleration of m1 and the tensions in the upper and lower portions of the string. Verify that your answers agree with part a if you set mp = 0. (Use any variable or symbol stated above along with the following as necessary: g.)
a=
Tupper=
Tlower=
The mass of the blue puck shown below is 40.0% greater than the mass of the green puck. Before colliding, the pucks approach each other with momenta of equal magnitudes and opposite directions, and the green puck has an initial speed of 10.0 m/s. Find the speeds the pucks have after the collision if half the kinetic energy of the system becomes internal energy during the collision.Explanation / Answer
a)
The net force accelerating m2 is = m2*g - T (with T = tension)
m2g - T = m2a --> T = m2g - m2a.
And the tension is also the force that accelerates m1:
T = m1a
set both equations equal to get
m2g - m2a = m1a --> solve for a:
a(m1 + m2) = m2g
a = m2g/(m1+m2)
T = m1*m2*g/(m1+m2)
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b)
T1 is the tension between m1 and the pulley
T2 is the tension between m2 and the pulley
the net force on m1 is T1 = m1a
the net force on m2 is m2g - T2 = m2a --> T2 = m2g - m2a
the net force on the pulley is T2 - T1
and the torque ? on the pulley is then = F*R = (T2 - T1)R
and the torque ? on the pulley is also ? = I*? = I*a/R (with I = moment of inertia of pulley, ? = angular acceleration of pulley)
--> (T2-T1)R = I*a/R --> divide by R
(T2 - T1) = Ia/R^2 plug in the expressions for T1 and T2:
(m2g - m2a) - m1a = Ia/R^2 --> solve for a:
a(-m2 - m1 - I/R^2) = - m2g
a = m2g/(m1+m2+I/R^2) with I = 1/2*mp*R^2
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plug a into the two equations for T1 and T2:
T1 = m1a
T2 = m2g-m2a
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