Question

A 65.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 480 kg · m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about a frictionless vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m/s relative to the Earth.

Answer 1

Answer:

A 65.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 480 kg · m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about a frictionless vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m/s relative to the Earth.

(a) In what direction and with what angular speed does the turntable rotate?

(b) How much work does the woman do to set herself and the turntable into motion?

The answers are

(a) 0.40625 rad/s ccw

(b) 112.73 J

Explanation:

To solve the question we have

Mass of woman = 65.  kg

Moment of inertia of the turntable = 480 kg·m²

Radius of turntable = 2.00 m

Speed of woman =  -1.50 m/s, clockwise motion

Sum of anguar momentum = constant

Therefore Inital total angular momentum = Finat toatal momentum

M₁ₐ, M₁ₓ = initial and final momentum of turn table and M₂ₐ, M₂ₓ for the woman

Therefore

M₁ₐ + M₂ₐ = M₁ₓ + M₂ₓ

Since the momentum = Inertia, I multiplied by angular velocity, ω, we have

I₁×ω₁ = -I₂×ω₂ and

ω₁ = -I₂×ω₂÷I₁

Which gives m·r²×v/r÷I₁ where ω = v/r and I = m·r²

=-m·r·v÷l₁, Substituting the values for the mass, radius, the speed and the moment of inertia of the turntable gives

=- 65×2×-1.5÷480 = 0.40625 rad/s ccw

The work done can be found from

Total work = change in kinetic ennergy

= Initial Kinetic energy of the system less final kinetic energy of the system

= Kinetic energy of the woman + Kinetic energy of the turntable - 0

1/2×I₁×ω₁² + 1/2×m×v₂²

= 1/2×480 kg·m²×(0.40625 rad/s)² +1/2×65 kg×(-1.5 m/s) ²

= 112.734375 kg·m/s ≈ 112.73 J

Answer 2

The angular velocity and work done is required.

The angular velocity of the turntable is 0.40625 rad/s counter clockwise.

The work done is 112.73 J.

$m_w$ = Mass of woman = 65 kg

$I_t$ = Moment of inertia of turntable = $480\ \text{kg/m}^2$

r = Radius = 2 m

v = Velocity = 1.5 m/s

Inertia of woman

$I_w=m_wr^2\\\Rightarrow I_w=65\times 2^2\\\Rightarrow I_w=260\ \text{kg/m}^2$

Angular velocity of woman

$\omega_w=\dfrac{v}{r}\\\Rightarrow \omega_w=\dfrac{1.5}{2}\\\Rightarrow \omega_w=0.75\ \text{rad/s}$

As the angular momentum is conserved

$I_w\omega_w=I_t\omega_t\\\Rightarrow \omega_t=\dfrac{I_w\omega_w}{I_t}\\\Rightarrow \omega_t=\dfrac{260\times 0.75}{480}\\\Rightarrow \omega_t=0.40625\ \text{rad/s}$

Work done is given by

$W=\dfrac{1}{2}m_wv^2+\dfrac{1}{2}I_t\omega_t^2\\\Rightarrow W=\dfrac{1}{2}\times 65\times 1.5^2+\dfrac{1}{2}\times 480\times 0.40625^2\\\Rightarrow W=112.73\ \text{J}$

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