Consider the rotating cylinder shown in Fig 5.26. Suppose that m= 5 kg, F = 0.60 N, and r = 0.20 m. Calculate (a) the torque acting on the cylinder, and (b) the angular acceleration of the cylinder.
(Moment of inertia of cylinder = 1212mr² 
(Ans: 0.12 N, 12 rad s⁻¹)






Data Given:


m = 5.0 kg

F = 0.60 N

r = 0.20 m

θ = 90°

To Find:


(a) Torque acting on the cylinder  = →T = ?

(b) Angular Acceleration  = ∝ = ?


Solution:


(a) Torque acting on the cylinder  = →T = ?

We know that 

→T = →r ｘ→F

or

→T = r F sin θ

By putting values

→T = 0.20 m ｘ 0.60 N ｘ sin 90°

→T = 0.12 N m ｘ 1

→T = 0.12 N m  -------Ans. (1)


(b) Angular Acceleration  = ∝ = ?

We know that 

T = I ∝

or

∝ = TI

where moment of inertia I= 12 mr², So

∝ = 2Tmr²

putting values

∝ = 2ｘ0.12Nm5.0kgｘ(0.20m)²

∝ = 0.24Nm5.0kgｘ0.04m²

∝ = 0.24Nm0.20kgm²

∝ = 1.20 rad s⁻²}`



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