12-15 A light string is wrapped around a solid cylinder that has a radius R = 5

Question

12-15 A light string is wrapped around a solid cylinder that has a radius R = 5 cm and a mass M = 12 kg. The cylinder is on a plane with an inclination angle of 32°. The free end of the string passes over an inertialess and frictionless pulley to a block with mass m = 2 kg as in the diagram at right. Find the linear acceleration of the cylinder and of the hang- ing block. [Hint: What is the relationship between the linear acceleration of the cylinder and that of the block?] M= 12 Kg R=5cm m= 2 kg 32° Problem 12-15

Explanation / Answer

Let the acceleration be a, tension be T.

By force equation on hanging block,

T - mg = ma

Torque equation

-T*R + fR= i*alpha where i=0.5MR^2 and alpha =a/R

so -T + f = 0.5M*a

also force equation on cylinder, Mg sin 32 degree -T-f= Ma

adding last two equations, -2T + Mg sin 32 degree = 1.5 Ma

or -T + 0.5*Mg sin 32 degree = 0.75 Ma

adding this to first equation,

-mg  + 0.5*Mg sin 32 degree = ma + 0.75 Ma

-2*9.8 + 0.5*12*9.8*sin 32 degree = (2+0.75*12) a

11a = 11.5592

a = 11.5592/11 = 1.05 m/s^2

acceleration of cylinder =  1.05 m/s^2 down the incline

acceleration of block =  1.05 m/s^2 upward

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