Question
A particle performs uniform circular motion with an angular momentum $L$. If the frequency of particle motion is doubled and its K.E. is halved, the angular momentum becomes:(a) $2 L$(b) $4 L$(c) $\frac{L}{2}$(d) $\frac{L}{4}$
Step 1
The frequency $f$ is related to the angular velocity by $f = \frac{\omega}{2\pi}$. Therefore, we can write $\omega = 2\pi f$. Let's call this equation (1). Show more…
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A particle performs uniform circular motion with an angular momentum $L$. If the frequency of a particle's motion is doubled and its kinetic energy is halved, the angular momentum becomes (a) $2 L$ (b) $4 L$ (c) $L / 2$ (d) $L / 4$
Rotational Motion
Round 1
A particle performing uniform circular motion has angular momentum $L$., its angular frequency is doubled and its $K . E$. halved, then the new angular momentum is $\{\mathrm{A}\} 1 / 2$ \{B $\} 1 / 4$ $\{\mathrm{C}\} 2 \mathrm{~L}$ $\{\mathrm{D}\} 4 \mathrm{~L}$
A particle performing uniform circular motion has angular frequency is doubled and its kinetic energy halved, then the new angular momentum is [2003] (A) $\frac{L}{4}$ (B) $2 L$ (C) $4 \mathrm{~L}$ (D) $\frac{L}{2}$
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