. Figure 3 shows a string is wrapped around a uniform solid pulley of radius R =
ID: 1483761 • Letter: #
Question
. Figure 3 shows a string is wrapped around a uniform solid pulley of radius R = 0.5m. The pulley can rotate freely about its axis. The loose end of the string is attached to a stone. The mass of the pulley mp = 2.5kg and the stone has a mass of ms = 1.5kg (different from the homework, the masses here are NOT the same). The moment of inertia for the pulley is Idisk = 1 2mpR2 . (a) What is the relationship between the magnitude of the angular acceleration ? and the the magnitude of the linear acceleration of the stone a? (b) Find the angular acceleration ? of the pulley as the stone descends. (c) Find the tension in the string.
2.50-kg pulley 1.50-kg stoneExplanation / Answer
a)
angular acceleration alpha = a/R
b)
net torque on pulley = T*R
but net torque = I*alpha ....(1)
along vertical
mg - T = m*a
T = mg - ma.........(2)
(mg - ma)*R = (/12)M*R^2*alpha
(mg - ma) = (1/2)M*a
(1.5*9.8)-(1.5*a) = (1/2)*2.5*a
linear acce;leration a = 5.345 m/s^2
alpha = a/R = 10.69 rad/s^2 <<_--------answer
++++++
c)
T = mg - ma
T = 1.5*(9.8-5.345) = 6.6825 N <<---answer
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