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Position and Linear Acceleration of a Point on a Rotating Wheel
(a) Position at $ t = 10 $ s:
We'll use the rotational kinematic equation to find the position:
\[ \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 \]
Since the wheel starts from rest, $ \omega_0 = 0 $ and $ \theta_0 = 0 $.
Substituting the given values:
\[ \theta = \frac{1}{2} \times 5.0 \times (10)^2 = 250 \, \text{rad} \]
Now, to find the position $ s $ at the bottom of the wheel, we can use the formula for arc length:
\[ s = r \theta \]
Given that the radius of the wheel $ r = 0.25 $ m, we have:
\[ s = 0.25 \times 250 = 62.5 \, \text{m} \]
So, the point that is initially at the bottom of the wheel at $ t = 10 $ s is $ 62.5 $ m from the center of the wheel.
(b) Linear Acceleration at $ t = 10 $ s:
We'll use the equation relating linear and angular acceleration:
\[ a_t = r \alpha \]
Given that the radius of the wheel $ r = 0.25 $ m and $ \alpha = 5.0 $ rad/s$^2$, we have:
\[ a_t = 0.25 \times 5.0 = 1.25 \, \text{m/s}^2 \]
So, the point's linear acceleration at $ t = 10 $ s is $ 1.25 $ m/s$^2$.