Ncert Exercises & Solutions

A mass of $5~\text {kg}$ is moving along a circular path of radius $1~\text {m}$. If the mas moves with $300~\text {rev/min}$, its kinetic energy would be

1. $250 \pi^2$
2. $100 \pi^2$
3. $5 \pi^2$
4. $0$

(a) Hint: Using angular velocity, we can find the linear speed.

Step 1: Find the angular speed of the mass.

$Given, mass = m = 5 kg$
$Radius = 1 m = R$
$Revolution per \min . ω = 300 rev / \min$
$= (300 \times 2 π) rad / \min$
$= (300 \times 2 \times 3 .14) rad / 60 s$
$= \frac{300 \times 2 \times 3 .14}{60} rad / s = 10 πrad / s$

Step 2: Find the kinetic energy of the mass.

$\begin{array}{l} ⇒ linear speed = v = ωR \\ = (\frac{300 \times 2 π}{60}) (1) \\ = 10 πm / s \\ KE = \frac{1}{2} {mv}^{2} \\ = \frac{1}{2} \times 5 \times (10 π)^{2} \\ = 100 π^{2} \times 5 \times \frac{1}{2} \\ = 250 π^{2} J \end{array}$

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