Question

1. Determine the moment of inertia of a 7.70 kgkg sphere of radius 0.779 mm when the axis of rotation is through its center.

2. A merry-go-round accelerates from rest to 0.70 rad/srad/s in 35 ss. Assuming the merry-go-round is a uniform disk of radius 8.0 mm and mass 3.20×10⁴ kgkg , calculate the net torque required to accelerate it.

3. A centrifuge rotor has a moment of inertia of 4.00×10⁻² kg⋅m2kg⋅m2 . How much energy is required to bring it from rest to 8270 rpm ?

I need help with these questions!

Answer 1

1) the moment of inertia of a solid sphere through its central axis is given as, I = 2/5*M.R²

Where M= mass of the sphere

= 7.70 kg

And R= radius of the solid sphere.

= 0.779 m

I= 2/5*M. R²

⁼ 2/5* 7.70* (0.779)²

= 1.87 kg. m²

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2) given  ₀ = angular velocity at time t =0 is also zero

₀ = 0 ; and  _t = 0.70 rad/s

and for t = 35 s

therefore,   , which is angular acceleration of merry when she goes around the disc and accelerates

   =   /t

= (_t -   ₀ ) / t

= 0.70 -0 / 35

= 0.02 rad/s²

and also torque of the acceleration can be found by the relation, = I.

so here I = moment of inertia of rotating disc = 1/2*M.R²

= 1/2 * 3.20 *10⁴ * 8²

= 1024000 kg.m²

so finally the  net torque required to accelerate it.

= I.

= 1024000 x 0.02

= 20480 N -m

= 20.48 KN-m