If the angular momentum of a rotating body is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The rotational kinetic energy $E$ is related to angular momentum $L$ and moment of inertia $I$ by the formula $E = \frac{L^2}{2I}$.Since $I$ remai

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(B) The rotational kinetic energy $E$ is related to angular momentum $L$ and moment of inertia $I$ by the formula $E = \frac{L^2}{2I}$.Since $I$ remains constant,we have $E \propto L^2$,which implies $\frac{E_2}{E_1} = \left( \frac{L_2}{L_1} ight)^2$.Given that the angular momentum is increased by $200\ \%$,the new angular momentum $L_2 = L_1 + 200\% 	ext{ of } L_1 = L_1 + 2L_1 = 3L_1$.Substituting this into the ratio: $\frac{E_2}{E_1} = \left( \frac{3L_1}{L_1} ight)^2 = 3^2 = 9$.Therefore,$E_2 = 9E_1$.The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} 	imes 100\% = \frac{9E_1 - E_1}{E_1} 	imes 100\% = 8 	imes 100\% = 800\%$.

If the angular momentum of a rotating body is increased by $200\ \%$,then its kinetic energy of rotation will be increased by .......... $\%$

Similar Questions

Two bodies have their moments of inertia $I$ and $2I$ respectively about their axis of rotation. If their kinetic energies of rotation are equal,their angular momentum will be in the ratio

$A$ molecule consists of two atoms each of mass $m$ and separated by a distance $d$. At room temperature,if the average rotational kinetic energy is $E$,then the angular frequency is:

Rotational kinetic energy of a given body about an axis is proportional to

If the rotational kinetic energy of an object is $50\%$ of its linear kinetic energy,then the object is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module