Question

A 3.0 kg bobbin consists of a central cylinder of radius 5cm and two end plates each of radius 6cm . It is placed on a slotted incline , where friction is sufficient to prevent sliding.A block of mass 4.5 kg is suspended from a cord wound around the bobbin and passing through the slot under the incline.If the bobbin is in static equilibrium, what is the angle of tilt of incline?

Answer 1

Let M is the mass of bobbin , R is the outer radius , r is the inner radius of the bobbin and $\theta$ is the inclination angle . when a block of mass m is suspended from a chord wound around the bobbin and passing through the slot under the incline. Torque will be increase as $(M + m)gsin\theta.R$

but since bobbin is in static equilibrium. so the two torque are equal and opposite direction.
e.g,. $(M+m)gsin\theta.R=mgr$

now, put the values of M, m, g, R and r .

so, (3 + 4.5) × 10 × sin$\theta$ × 6= 4.5 × 10 × 5cm
7.5 × 10 × sin$\theta$ × 6 = 45 × 5

$sin\theta$ = 45 × 5/(75 × 6) = 1/2

$\theta=30^{\circ}$

Answer 2

Here is the answer for you!!