An old 78 rpm record rotates through an angle of $320 ^ { \circ }$ as it slows down uniformly from 78.0 rpm to 22.8 rpm. What is the magnitude of the angular acceleration of the record?
A) $5.46 \mathrm { rad } / \mathrm { s } ^ { 2 }$
B) $2.34 \mathrm { rad } / \mathrm { s } ^ { 2 }$
C) $10.9 \mathrm { rad } / \mathrm { s } 2$
D) $8.35 \mathrm { rad } / \mathrm { s } ^ { 2 }$
E) $6.50 \mathrm { rad } / \mathrm { s } ^ { 2 }$

Question

An old  78 rpm record rotates through an angle of $320 ^ { \circ }$ as it slows down uniformly from  78.0 rpm  to  22.8 rpm. What is the magnitude of the angular acceleration of the record?
A) $5.46 \mathrm { rad } / \mathrm { s } ^ { 2 }$
B) $2.34 \mathrm { rad } / \mathrm { s } ^ { 2 }$
C) $10.9 \mathrm { rad } / \mathrm { s } 2$
D) $8.35 \mathrm { rad } / \mathrm { s } ^ { 2 }$
E) $6.50 \mathrm { rad } / \mathrm { s } ^ { 2 }$

Quizplus · Accepted Answer

The angular acceleration is given by:$$\alpha = \frac{\omega_f - \omega_i}{t}$$where $\omega_i$ is the initial angular velocity, $\omega_f$ is the final angular velocity, and $t$ is the time.We are given that $\omega_i = 78.0 \mathrm { rpm }$, $\omega_f = 22.8 \mathrm { rpm }$, and $\theta = 320 ^ { \circ } = 5.58 \mathrm { rad }$. We can convert the angular velocities to radians per second:$$\omega_i = 78.0 \mathrm { rpm } \left( \frac{2\pi \mathrm { rad }}{60 \mathrm { s }} \right) = 8.17 \mathrm { rad } / \mathrm { s }$$$$\omega_f = 22.8 \mathrm { rpm } \left( \frac{2\pi \mathrm { rad }}{60 \mathrm { s }} \right) = 2.40 \mathrm { rad } / \mathrm { s }$$We do not know the time $t$, but we can use the fact that the record rotates through an angle of $5.58 \mathrm { rad }$ to find it:$$\theta = \omega_i t + \frac{1}{2} \alpha t^2$$Since the record is slowing down, the angular acceleration is negative. We can solve for $t$:$$t = \frac{\omega_f - \omega_i}{\alpha}$$Substituting this into the equation for $\theta$, we get:$$\theta = \omega_i \left( \frac{\omega_f - \omega_i}{\alpha} \right) + \frac{1}{2} \alpha \left( \frac{\omega_f - \omega_i}{\alpha} \right)^2$$Simplifying, we get:$$\theta = \frac{(\omega_f - \omega_i)^2}{2\alpha}$$Solving for $\alpha$, we get:$$\alpha = \frac{(\omega_f - \omega_i)^2}{2\theta}$$Substituting the values we know, we get:$$\alpha = \frac{(2.40 \mathrm { rad } / \mathrm { s } - 8.17 \mathrm { rad } / \mathrm { s })^2}{2(5.58 \mathrm { rad })} = 5.46 \mathrm { rad } / \mathrm { s } ^ { 2 }$$Therefore, the magnitude of the angular acceleration of the record is $5.46 \mathrm { rad } / \mathrm { s } ^ { 2 }$.

An Old 78 Rpm Record Rotates Through an Angle 320∘320 ^ { \circ }320∘

An Old 78 Rpm Record Rotates Through an Angle $320 ^ { \circ }$