The mass m moves in a circular arc of angular amplitude 60° . The minimum coefficient of friction between 4m and surface of table to prevent slipping is,

Question

The mass m moves in a circular arc of angular amplitude 60° . The minimum coefficient of friction between 4m and surface of table to prevent slipping is, <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-zmsxcubgw8oss6eg.jpg" /> (A) 0.25 (B) 0.45 (C) 0.5 (D) 0.2

Tardigrade Admin · Accepted Answer

For 60° circular arc, mg(ℓ - ℓ cos 60 ° )=(1/2)mvmax2 mg=(m vmax2/ℓ ) ...(i) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-zmsxcubgw8os2uae.jpg" /> Maximum tension in the rope =mg+(m vmax2/ℓ )=2mg . For minimum friction coefficient, μ 4mg=2mg μ =0.5

Q. The mass $m$ moves in a circular arc of angular amplitude $6 0^{\circ}$ . The minimum coefficient of friction between $4 m$ and surface of table to prevent slipping is,

$0.25$

$0.45$

$0.5$

$0.2$

For $6 0^{\circ}$ circular arc,
$m g (ℓ - ℓ cos 6 0^{\circ}) = \frac{1}{2} m v_{ma x}^{2}$
$m g = \frac{m v _{ma x}^{2}}{ℓ}$ ...(i)

Maximum tension in the rope $= m g + \frac{m v _{ma x}^{2}}{ℓ} = 2 m g$ .
For minimum friction coefficient,
$μ 4 m g = 2 m g$
$μ = 0.5$

Q. The mass m moves in a circular arc of angular amplitude 60∘ . The minimum coefficient of friction between 4m and surface of table to prevent slipping is,

0.25

0.45

0.5

0.2