Question

A uniform drum of radius R and mass M rolls without slipping down a plane inclined at angle . Find its acceleration along the plane (translational acceleration). The moment of inertia of the drum about its axis through the center is I = MR^2/2 .

Answer 1

here,

assume l as the lenght of the incline
h = l*sin (theta) is the heght of the incline

kinetic energy at the bottom Ek = 1/2m*V^2+1/4m*V^2 = 3m*V^2/4

initial potential energy at the top E = m*g*l*sin (theta)

energies equate :

3m*V^2/4 = m*g*l*sin (theta)

mass m cross

3V^2 = 4*g*l*sin (theta)

V^2 = 4*g*l*sin (theta)/3...

final speed at the bottom

V^2 = 2*a*l

4*g*l*sin (theta)/3.= 2*a*l

lenght l cross

4*g*sin (theta) = 6a

a = 4*g*sin (theta)/6

a = 6.54*sin (theta)

the translation accelration is 6.54 * sin(theta)