Acable passes over a pulley. because the cable grips the pulley and the pulley has non- zero mass, the tension in the cable is not the same on opposite sides of the pulley. the force on one side is 167 n, and the force on the other side is 42 n. assuming that the pulley is a uniform disk of mass 1.06 kg and radius 0.433 m, find the magnitude of its angular acceleration. for a uniformi disk, i = (1/2) m,2.1 answer in units of rad/s





544.68604 rad/s²


m = Mass of disk = 1.06 kg

R = Radius of disk = 0.433 m

T = Tension

T_2 = 167 N

T_1 = 42 N

Moment of inertia is given by

I=\dfrac{1}{2}mR^2\\\Rightarrow I=\dfrac{1}{2}\times 1.06\times 0.433^2

The resultant torque of the system will be given by

(T_2-T_1)R=\tau\\\Rightarrow (T_2-T_1)R=I\alpha\\\Rightarrow \alpha=\dfrac{(T_2-T_1)R}{I}\\\Rightarrow \alpha=\dfrac{(167-42)\times 0.433}{\dfrac{1}{2}\times 1.06\times 0.433^2}\\\Rightarrow \alpha=544.68604\ rad/s^2

The angular acceleration of the disk is 544.68604 rad/s²

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