A uniform cylinder of mass m1 and radius R is pivoted on frictionless bearings.

Question

A uniform cylinder of mass m1 and radius R is pivoted on frictionless bearings. A mass less string wrapped around the cylinder is connected to a block of mass m2 that is on a frictionless incline of angle theta as shown below. The system is released from rest when the block is at a vertical distance h above the bottom of the incline. (Use any variable or symbol stated above along with the following as necessary: g.) What is the acceleration of the block? a = What is the tension in the string? T = What is the speed of the block as it reaches the bottom of the incline? v = Evaluate your answers for the special case where theta = 90degree and m1 = 0. Are your answers what you would expect for this special case? a = m/s2 T = N v = m/s

Explanation / Answer

a)

m2*gSin - T = m2*a

Torque = T*R = I = (1/2*m1*R^2)*(a/R)

So, T = 1/2*m1*a

So, m2*gSin - 1/2*m1*a = m2*a

a = m2*g*Sin/(m2 + 0.5*m1)

b)

T = 1/2*m1*a = 0.5*m1*m2*g*Sin/(m2 + 0.5*m1)

c)

Distance s = h/Sin

Using v^2 = u^2 + 2as,

v^2 = 0 + 2*[m2*g*Sin/(m2 + 0.5*m1)]*(h/Sin)

v = [2*m2*g*h/(m2 + 0.5*m1)]^0.5

d)

Putting m1 = 0 and = 90 deg, we get

a = m2*g*Sin90/(m2 + 0.5*0) = g (Free fall)

T = 0.5*0*m2*g*Sin90/(m2 + 0.5*m1) = 0 (No tension)

v =  [2*m2*g*h/(m2 + 0.5*0)]^0.5 = (2gh)

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