Question

1. A uniform cylinder of mass m and radius r is rolling down a slope of inclination 30°. The cylinder rolls without slipping. At what rate does the cylinder accelerate down the slope?

Answer 1

Here we need to find rate at which cylinder is accelerating down the slope, So

We know that relation between linear acceleration and angular acceleration is given by:

a = *r

Now torque applied on cylinder will be given by:

= I*, So

= angular acceleration = /I

= torque applied due to weight force along the incline = rxF = r*F*sin

F = Weight force = m*g

= m*g*r*sin

I = moment of inertia of solid cylinder rolling down the slope (Using parallel axis theorem)

I = I0 + m*r^2 = (1/2)*m*r^2 + m*r^2 = (3/2)*m*r^2

I0 = moment of inertia of cylinder rotating about central axis = (1/2)*m*r^2

So,

= (m*g*r*sin )/(3*m*r^2/2) = 2*m*g*r*sin /(3*m*r^2)

= 2*g*sin /(3*r)

Now linear acceleration will be:

a = *r

a = (2*g*sin /(3*r))*r

a = (2/3)*g*sin

Since = 30 deg, So

a = (2/3)*9.8*sin 30 deg

a = 3.27 m/s^2

Let me know if you've any query.