Question

An electric motor rotating a workshop grinding wheel at $1.00\times {10}^2 rev/\min$is switched off. Assume the wheel has a constant negative angular acceleration of magnitude$2.00 rad/s^2$. (a) How long does it take the grinding wheel to stop? (b) Through how many radians has the wheel turned during the time interval found in part (a)?

Short answer

The solution for the angular acceleration of magnitude of time taken and the radians are

(a) t = 5.24 s

(b)$\theta =27.4rad$

Step-by-step solution

Convert the given units and derive angular speed

First converting units:

Multiply the angular speed$\left({\omega }_i\right)$ by a conversion factor to convert its units from$(rev/\min )to(rad/s)$

${\omega }_i=1\times {10}^2\left(\frac{rev}{\min }\right)\left(\frac{2\pi rad}{1rev}\right)\left(\frac{1\min }{60s}\right)=\frac{10\pi }{3}rad/s$

Second: solving the problem:

Model the electric motor as a rigid object under constant angular acceleration and use the following equation:

role="math" localid="1663710369265" ${\mathit{\omega }}_{\mathbf{f}}\mathbf{=}{\mathit{\omega }}_{\mathbf{i}}\mathbf{+}\mathit{\alpha }\mathit{l}$

(a)

Solve for (t):

$t=\frac{{\omega }_f-{\omega }_i}{\alpha }$

Substitute numerical values:

$$\begin{gathered} t=\frac{0-\frac{10\pi }{3}}{(-2)} \\ =5.24 s \end{gathered}$$

Hence, the answer is 5.24 s.

Derive using constant angular acceleration

(b)

Similarly, from rigid object under constant angular acceleration model, use the following equation:

${\omega }_f^2={\omega }_i^2+2\alpha \theta$

Solve for

$\theta =\frac{{\omega }_f^2-{\omega }_i^2}{2\alpha }$

Substitute numerical values:

$$\begin{gathered} \theta =\frac{0-{\left(\frac{10\pi }{3}\right)}^2}{2(-2)} \\ =27.4 rad \end{gathered}$$

Hence, the answer is 27.4 rad.