A rolling wheel of diameter of 68 cm slows down uniformly from 8.4 m/s to rest over a distance of 115 m. What is the magnitude of its angular acceleration if there was no slipping?
A) $0.90 \mathrm { rad } / \mathrm { s } ^ { 2 }$
B) $5.7 \mathrm { rad } / \mathrm { s } ^ { 2 }$
C) $11 \mathrm { rad } / \mathrm { s } ^ { 2 }$
D) $1.8 \mathrm { rad } / \mathrm { s } ^ { 2 }$

Question

A rolling wheel of diameter of 68 cm slows down uniformly from 8.4 m/s to rest over a distance of 115 m. What is the magnitude of its angular acceleration if there was no slipping? 
A) $0.90 \mathrm { rad } / \mathrm { s } ^ { 2 }$
B) $5.7 \mathrm { rad } / \mathrm { s } ^ { 2 }$
C) $11 \mathrm { rad } / \mathrm { s } ^ { 2 }$
D) $1.8 \mathrm { rad } / \mathrm { s } ^ { 2 }$

Quizplus · Accepted Answer

The angular acceleration can be found using the relation between linear acceleration and angular acceleration, $a = \alpha r$, where $a$ is the linear acceleration, $\alpha$ is the angular acceleration, and $r$ is the radius of the wheel. First, we find the linear acceleration using the equation $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m/s), $u$ is the initial velocity (8.4 m/s), $a$ is the acceleration, and $s$ is the distance (115 m). Solving for $a$, we get $a = -u^2 / (2s) = -(8.4)^2 / (2 \times 115) = -0.306 \, \text{m/s}^2$. The negative sign indicates deceleration. The radius of the wheel is half the diameter, so $r = 68 / 2 = 34 \, \text{cm} = 0.34 \, \text{m}$. Thus, $\alpha = a / r = -0.306 / 0.34 \approx -0.90 \, \text{rad/s}^2$. The magnitude of angular acceleration is $0.90 \, \text{rad/s}^2$, ignoring the negative sign which indicates direction.

A Rolling Wheel of Diameter of 68 Cm Slows Down 0.90rad/s20.90 \mathrm { rad } / \mathrm { s } ^ { 2 }0.90rad/s2

A Rolling Wheel of Diameter of 68 Cm Slows Down $0.90 \mathrm { rad } / \mathrm { s } ^ { 2 }$