A semicircular disc of radius R and mass M is pulled by a horizontal force F so that it moves with uniform velocity. The coefficient of friction between disc and ground is μ=(3/3 π). The value of θ in degree is .

Question

A semicircular disc of radius R and mass M is pulled by a horizontal force F so that it moves with uniform velocity. The coefficient of friction between disc and ground is μ=(3/3 π). The value of θ in degree is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As the disc moves with uniform velocity

F = μ m g

and

N = m g

The force

F

and friction force will form a couple.
Balancing torque about centre of mass

μ m g R (1 - sin θ) = m g \frac{4 R}{3 π} sin θ

\Rightarrow sin θ = \frac{1}{2}

\Rightarrow θ = 3 0^{\circ}

Q. A semicircular disc of radius R and mass M is pulled by a horizontal force F so that it moves with uniform velocity. The coefficient of friction between disc and ground is μ=3π3​. The value of θ in degree is ____.

As the disc moves with uniform velocity F=μmg and N=mg The force F and friction force will form a couple. Balancing torque about centre of mass μmgR(1−sinθ)=mg3π4R​sinθ ⇒sinθ=21​ ⇒θ=30∘

Q. A semicircular disc of radius $R$ and mass $M$ is pulled by a horizontal force $F$ so that it moves with uniform velocity. The coefficient of friction between disc and ground is $μ = \frac{3}{3 π}$ . The value of $θ$ in degree is ____.

As the disc moves with uniform velocity $F = μ m g$ and $N = m g$
The force $F$ and friction force will form a couple.
Balancing torque about centre of mass
$μ m g R (1 - sin θ) = m g \frac{4 R}{3 π} sin θ$

$\Rightarrow sin θ = \frac{1}{2}$
$\Rightarrow θ = 3 0^{\circ}$