If earth were to suddenly contract to remain two third of its present size

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 brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m2. 

(a) 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?

(a) Moment of inertia of the man-platform system = 7.6 kg m2

Moment of inertia when the man stretches his hands to a distance of 90 cm,

2 × m r2 = 2 × 5 × (0.9)2 

            = 8.1 kg m2 

Initial moment of inertia of the system, Ii2 

Angular speed, ωi = 300 rev/min 

Angular momentum, Li = Iiωi  =  15.7 × 30                         ...(i) 

Moment of inertia when the man folds his hands to a distance of 20 cm

2 × mr2 = 2 × 5 (0.2)22 

Final moment of inertia, If = 7.6 + 0.4 = 8 kg m2  

Final angular speed = ωf 

Final angular momentum, Lf = Ifωf = 0.79 ωf                   .... (ii) 

From the conservation of angular momentum, we have

                       Iiωi  =  Ifωf 

ωf =

 

(b) 

Kinetic energy is not conserved in the given process. But, the kinetic energy increases as there is a decrease in the moment of inertia. 

The additional kinetic energy comes from the work done by the man to fold his hands toward himself.

Tardigrade - CET NEET JEE Exam App

© 2022 Tardigrade®. All rights reserved

Text Solution

24/n hours24 n hours`24//n^(2)` hours`24n^(2)` hours

Answer : C

Solution : Using principle of conservation of angular momentum, <br> `I_(1)omega_(1)=I_(2)omega_(2)` <br> `2/5MR^(2)xx(2pi)/24=2/5M(R/n)^(2)xx(2pi)/TrArrT=24/(n^(2))` hour

Text Solution

Solution : Here, `R_(2) = (1)/(3)R_(1), T_(1) = 24 hrs, T_(2) = ?` <br> According to the principle of conservation of angular momentum <br> `I_(2)omega_(2) = I_(1)omega_(1)` <br> `I_(2) ((2pi)/(T_(2))) = I_(1) ((2pi)/(T_(1)))` <br> `T_(2) = (I_(2))/(I_(1)) xx T_(1) = ((2)/(5)MR_(2)^(2))/((2)/(5)MR_(1)^(2)) xx T_(1)` <br> `T_(2) = T_(1) xx ((R_(2))/(R_(1)))^(2)` <br> `= 24 xx ((1)/(3))^(2) = 2 hr 40 min`. <br> `:. T_(1) - T_(2) = 24 hr - 2hr 40 min` <br> `= 21 hr 20 min`

Answer

Verified

We are given that the Earth is contracted to half its present size. Change in size implies change in radius, so, here the present radius of earth $\left( {{R}_{1}} \right)$ is reduced by half. That is, after contraction the radius becomes,${{R}_{2}}=\dfrac{{{R}_{1}}}{2}$ ………………………………. (1)But during the contraction, the mass remains constant. That is,${{M}_{1}}={{M}_{2}}=M$ ……………………………… (2)Due to the above mentioned contraction, the angular velocity of Earth’s rotation changes. Also, we know that angular velocity is given by,$\omega =\dfrac{2\pi }{T}$ Angular velocity before contraction,$\Rightarrow {{\omega }_{1}}=\dfrac{2\pi }{{{T}_{1}}}$ ………………………… (3)Angular velocity after contraction,$\Rightarrow {{\omega }_{2}}=\dfrac{2\pi }{{{T}_{2}}}$ ……………………………….. (4)There is another quantity that is undergoing change due to contraction of earth – moment of inertia.Moment of inertia of a solid sphere is given by,$I=\dfrac{2}{5}M{{R}^{2}}$ Moment of inertia before contraction,${{I}_{1}}=\dfrac{2}{5}M{{R}_{1}}^{2}$ …………………………. (5)Moment of inertia after contraction,${{I}_{2}}=\dfrac{2}{5}M{{R}_{2}}^{2}=\dfrac{2}{5}M{{\left( \dfrac{{{R}_{1}}}{2} \right)}^{2}}=\dfrac{{{I}_{1}}}{4}$ ………………………………. (6)Now, by the law of conservation of angular momentum, $\dfrac{dL}{dt}=0$ $\Rightarrow L=I\omega $= constant$\Rightarrow {{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}$ …………………………….. (7)Substituting (3), (4) and (6) in (7), we get,${{I}_{1}}\left( \dfrac{2\pi }{{{T}_{1}}} \right)=\left( \dfrac{{{I}_{1}}}{4} \right)\left( \dfrac{2\pi }{{{T}_{2}}} \right)$ $\Rightarrow {{T}_{2}}=\dfrac{{{T}_{1}}}{4}$ ………………….. (8)But we know the present time period rotation of our planet Earth as 24 hours. Therefore, ${{T}_{1}}=24Hrs$ …………… (9) Substituting (9) in equation (8), we get,${{T}_{2}}=\dfrac{24}{4}$ $\Rightarrow {{T}_{2}}=6Hrs$ Time period of rotation of Earth is what we call a day. So, for an Earth that is contracted to half its present size with mass constant, the duration of a day will be just 6hrs.

So, the correct answer is “Option C”.

Note: Though we have discussed the case where the size of the Earth is reduced by half, we could actually generalize the above relation. Let us consider that Earth is contracted to $\dfrac{1}{n}th$ of its present size,

$\Rightarrow {{R}_{2}}=\dfrac{{{R}_{1}}}{n}$ $\Rightarrow {{I}_{2}}=\dfrac{2}{5}M{{\left( \dfrac{{{R}_{1}}}{n} \right)}^{2}}=\dfrac{{{I}_{1}}}{{{n}^{2}}}$ Substituting in (7) and then rearranging,${{T}_{2}}=\dfrac{{{T}_{1}}}{{{n}^{2}}}=\dfrac{24}{{{n}^{2}}}$ This generalization is under the condition that the mass of earth remains constant.