Question

how long does it take for a rotating object to speed up from 15.0 rad/s to 33.3 rad/s if it has a uniform angular acceleration of 3.45 rad/s2?
a) 5.30 s
b) 9.57 s
c) 6.31 s
d) 10.6 s
e) 53.0 s
f) 63.1 s

Explanation:

Step1: Recall angular motion formula

We use the angular motion formula \( \omega_f = \omega_i + \alpha t \), where \( \omega_f \) is final angular velocity, \( \omega_i \) is initial angular velocity, \( \alpha \) is angular acceleration, and \( t \) is time. Rearranging for \( t \), we get \( t=\frac{\omega_f - \omega_i}{\alpha} \).

Step2: Substitute values

Given \( \omega_i = 15.0 \, \text{rad/s} \), \( \omega_f = 33.3 \, \text{rad/s} \), \( \alpha = 3.45 \, \text{rad/s}^2 \). Substitute into the formula: \( t=\frac{33.3 - 15.0}{3.45}=\frac{18.3}{3.45} \approx 5.30 \, \text{s} \)? Wait, no, wait, 33.3 - 15 is 18.3? Wait 33.3 - 15.0 is 18.3? Wait 33.3 - 15 = 18.3? Wait 33.3 - 15.0 is 18.3? Then 18.3 divided by 3.45: 3.45*5 = 17.25, 18.3 -17.25=1.05, 1.05/3.45≈0.304, so total t≈5.304 s, which is ~5.30 s. Wait but wait, did I miscalculate? Wait 33.3 -15 is 18.3? Wait 33.3 -15.0 = 18.3. Then 18.3 /3.45: let's do 18.3 ÷3.45. Multiply numerator and denominator by 100: 1830 ÷345. Divide numerator and denominator by 15: 122 ÷23 ≈5.304. So t≈5.30 s. Wait but the option a is 5.30 s. Wait but wait, maybe I made a mistake? Wait no, let's check again. Wait the formula is correct: \( \omega_f = \omega_i + \alpha t \), so solving for t: \( t = (\omega_f - \omega_i)/\alpha \). So (33.3 -15.0)/3.45 = 18.3/3.45 = 5.304...≈5.30 s. So the correct answer should be a) 5.30 s. Wait but wait, maybe I misread the numbers? Let me check again. Initial angular velocity \( \omega_i = 15.0 \, \text{rad/s} \), final \( \omega_f = 33.3 \, \text{rad/s} \), acceleration \( \alpha = 3.45 \, \text{rad/s}^2 \). So difference in velocity: 33.3 -15 = 18.3. Divide by 3.45: 18.3 /3.45 = 5.304, which is approximately 5.30 s. So option a.

Answer:

a) 5.30 s