Question

A man stands on a rotating platform with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then bring his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m²a. What is his new angular speed? (Neglect friction)b. Is kinetic energy conserved in the process? If not from where does the change come about?

Answer 1

(a)
It is given that, Moment of Inertia of the man-platform system = 7.6 Kgm2
Moment of inertia when the man stretches his hands to a distance of 90 cm.
We know, I = 2 × mr²
Where, m is mass and r is distance of mass from axis of rotation.

Using the above formula, we get,
⇒ I = 2 × 5Kg × (0.9m)²
⇒ I = 8.1 Kgm²

Now, the Initial moment of inertia of the system,
Ii = 7.6 + 8.1 = 15.7 Kgm²

Initial angular speed, ωi = 30 rev/min = 30 × 2π/60 = π rad/s
Angular momentum, Li = Iiωi
Li = 15.7Kgm2 × π rad/s
⇒ Li = 49.298 Kgm² /sec ...…(i)

Moment of inertia when the man folds his hands to a distance of 20 cm,
I = 2 × mr²
⇒ I = 2 × 5Kg × (0.2m)²
⇒ I = 0.4 Kgm²
Final moment of inertia, If= 7.6 + 0.4 = 8 Kgm²
Final angular speed = ωf
Final angular momentum , Lf = Ifωf
Lf = 0.79 ωf ....…(ii)

From the conservation of angular momentum, We have
Iiωi = Ifωf
From equation (i) and (ii),
ωf = {49.298kgm²/sec}/8 Kgm²
ωf = 6.16225 rad/s

(b) Kinetic energy is not conserved in the given process. In fact, with the decrease in moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands towards himself.

Answer 2

Answer:

Let assume that , position vector of ith particle with respect to origin is r_ir

i

A man stands on a rotating platform with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then bring his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m²a. What is his new angular speed? (Neglect friction)b. Is kinetic energy conserved in the process? If not from where does the change

, position vector of ith particle with respect to centre of mass is r'_ir

i

′

and position vector of centre of mass with respect to origin is R.

It is given that, r'_i=r_i-Rr

i

′

=r

i

−R

so, r_i=r'_i+Rr

i

=r

i

′

+R

It is also given, that,

p_i=p'_i+m_iVp

i

=p

i

′

+m

i

V

Taking cross product of this relation with r_ir

i

,

\begin{gathered}r_i\times p_i=r_i\times p'_i+r'_i\times m_iV\\r_i\times p_i=(r'_i+R)\times p'_i+(r'_i+R)\times m_iV\\r_i\times p_i=r'_i\times p'_i+R\times p'_i+r'_i\times m_iV+R\times m_iV\\L=L'+R\times p'_i+r'_i\times m_iV+R\times m_iV\end{gathered}

r

i

×p

i

=r

i

×p

i

′

+r

i

′

×m

i

V

r

i

×p

i

=(r

i

′

+R)×p

i

′

+(r

i

′

+R)×m

i

V

r

i

×p

i

=r

i

′

×p

i

′

+R×p

i

′

+r

i

′

×m

i

V+R×m

i

V

L=L

′

+R×p

i

′

+r

i

′

×m

i

V+R×m

i

V

where, R\times p'_i=0R×p

i

′

=0

And r'_i m_iV=0r

i

′

m

i

V=0

So, L=L'+R\times m_iVL=L

′

+R×m

i

V

In simply you can say that, L = L' + R × MV