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 = `\frac {1}{2}`mr² 
(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  = `\vec T` = ?

(b) Angular Acceleration  = ∝ = ?


Solution:


(a) Torque acting on the cylinder  = `\vec T` = ?

We know that 

`\vec T` = `\vec r` ｘ`\vec F`

or

`\vec T` = r F sin Î¸

By putting values

`\vec T` = 0.20 m ï½˜ 0.60 N ï½˜ sin 90°

`\vec T` = 0.12 N m ï½˜ 1

`\vec T` = 0.12 N m  -------Ans. (1)


(b) Angular Acceleration  = ∝ = ?

We know that 

T = `\I` ∝

or

∝ = `\frac {T}{I}`

where moment of inertia `\I`= `\frac {1}{2}` mr², So

∝ = `\frac {2T}{mr²}`

putting values

∝ = `\frac {2ｘ0.12 Nm}{5.0 kg ｘ(0.20 m)²}`

∝ = `\frac {0.24 Nm}{5.0 kg ｘ0.04 m²}`

∝ = `\frac {0.24 Nm}{0.20 kg m²}`

∝ = 1.20 rad s⁻²}`



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