In a certain cyclotron, a proton of mass $1.67 \times 10 ^ { - 27 } \mathrm {~kg}$ moves in a circle of diameter 1.6 m with an angular speed of $2.0 \times 10 ^ { 6 } \mathrm { rad } / \mathrm { s }$ what is the angular momentum of the proton?
A) $1.8 \times 10 ^ { - 21 } \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } / \mathrm { s }$
B) $3.2 \times 10^{-21} \mathrm{~kg} \cdot \mathrm{m}^{2 / \mathrm{s}}$
C) $1.3 \times 10 ^ { - 2 } \mathrm {~kg} \cdot \mathrm { m } 2 / \mathrm { s }$
D) $2.1 \times 10 ^ { - 21 } \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } / \mathrm { s }$

Question

Quizplus · Accepted Answer

Angular momentum is given by $L = I \omega$, where $I$ is the moment of inertia and $\omega$ is the angular speed. For a point mass rotating in a circle, the moment of inertia is $I = mr^2$, where $m$ is the mass and $r$ is the radius of the circle. Substituting these expressions into the equation for angular momentum, we get $L = mvr$. Plugging in the given values, we get $L = (1.67 \times 10^{-27} \mathrm{kg})(0.8 \mathrm{m})(2.0 \times 10^6 \mathrm{rad/s}) = 2.1 \times 10^{-21} \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}$.

In a Certain Cyclotron, a Proton of Mass 1.67×10−27 kg1.67 \times 10 ^ { - 27 } \mathrm {~kg}1.67×10−27 kg

In a Certain Cyclotron, a Proton of Mass $1.67 \times 10 ^ { - 27 } \mathrm {~kg}$