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Question

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
 


Solution

Let m and r be the respective masses of the hollow cylinder and the solid sphere. 
The moment of inertia of the hollow cylinder about its standard axis, II=mr2
The moment of inertia of the solid sphere about an axis passing through its centre,
III=25mr2
We have the relation:
τ=I α
Where, α = Angular acceleration

τ = Torque
I = Moment of inertia
For the hollow cylinder, τI=IIαI
For the solid sphere, For the solid sphere, τII=IIIαII
As an equal torque is applied to both the bodies, τ1=τ2
 αIIαI=IIIII=mr225mr2=52αII>αI     (i)
Now using the relation:
ω=ω0+α t
Where
ω0 =Initial angular velocity
T= Time of rotation
ω = Final angular velocity
For equal ω0 and t, we have:
ω  α(ii)
From equations (i) and (ii), we can write:
ωII>ωI
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.


Physics
NCERT Textbook
Standard XI

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