If `M` is mass and `R` is radius of the hollow cylinder and the solid sphere, then

`M.I.` of hollow cylinder about its axis of symmetry, `I_(1) = MR^(2)`, and

`M.I.` of solid sphere about an axis through its centre, `I_(2) = (2)/(5)MR^(2)`

Torque applied, `tau = I_(1) alpha_(1) = I_(2) alpha_(2) :. (alpha_(2))/(alpha_(1)) = (I_(1))/(I_(2)) = (MR^(2))/((2)/(5) MR^(2)) = (5)/(2) :. alpha_(2) gt alpha_(a)`

From `omega = omega_(0) + alpha t`, we find that for given `omega_(0) and t, omega_(2) gt omega_(1)` i.e. angular speed of solid sphere will be greater than the angular speed of hollow cylinder.

Shanti Roy

Asked: 2 years ago2022-10-29T02:53:23+05:30
2022-10-29T02:53:23+05:30In: General Awareness

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

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Let M be the mass and R the radius of the hollow cylinder, and also of the solid sphere. Their moments of inertia about the respective axes are `l_(1) =MR^(2) ” and ” l_(2)=2//5 MR^(2)`

Let T be the magnitude of the torque applied to the cylinder and the sphere, producing angular accelerations `alpha_(1) ” and ” alpha_(2)` respectively. Then `T=l_(1) alpha_(1)=l_(2) alpha_(2)`

The angular acceleration 04 produced in the sphere is larger. Hence, the sphere will acquire larger angular speed after a given time.