A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) :

Question

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) : (A) 12 rad/s2 (B) 16 rad/s2 (C) 10 rad/s2 (D) 20 rad/s2

Tardigrade Admin · Accepted Answer

40 + f = m(R α) .....(i) 40 × R - f × R = mR2 α 40 - f = mR α ...... (ii) From (i) and (ii) α = (40/mR) = 16

$12 r a d / s^{2}$

$16 r a d / s^{2}$

$10 r a d / s^{2}$

$20 r a d / s^{2}$

$40 + f = m (R α)$ .....(i)
$40 \times R - f \times R = m R^{2} α$
$40 - f = m R α$ ...... (ii)
From (i) and (ii)
$α = \frac{40}{m R} = 16$

12rad/s2

16rad/s2

10rad/s2

20rad/s2

40+f=m(Rα) .....(i) 40×R−f×R=mR2α 40−f=mRα ...... (ii) From (i) and (ii) α=mR40​=16

$12 r a d / s^{2}$

$16 r a d / s^{2}$

$10 r a d / s^{2}$

$20 r a d / s^{2}$

$40 + f = m (R α)$ .....(i)
$40 \times R - f \times R = m R^{2} α$
$40 - f = m R α$ ...... (ii)
From (i) and (ii)
$α = \frac{40}{m R} = 16$