Question

A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. When the power to the grinder is turned off, the grinding wheel slows with constant angular acceleration and takes 80.0 s to come to a rest. (a) What was the angular acceleration (in rad/s2) of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive?

Answer 1

Answer:

Angular acceleration = -1.57 Rad/s^2

Number of revolutions = 800 revolutions

Explanation:

Mass = 1.15kg

diameter of the grinding wheel = 22.0cm

angular speed = 20 rev/s

1 Rev/s = 6.283 rad/s

20 Rev/s = 20 * 6.283 = 125.664 rad/s

Angular acceleration = Angular speed/ time taken for for the wheel rotation

Angular accleration = 125.664/80

Angular acceleration = -1.57 rad/s^2 (since the wheel is rotating counterclockwise)

number of revs = 0.5∝t²

number of revs = 0.5 * (1.57/2π) * 80²

number of revs = 800 revolutions

Answer 2

Complete Question:

A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. When the power to the grinder is turned off, the grinding wheel slows with constant angular acceleration and takes 80.0 s to come to a rest.

(a) What was the angular acceleration (in rad/s2) of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive?

(b) How many revolutions did the wheel make during the time it was coming to rest?

Answer:

(a) the angular acceleration is -1.571 rad/s²

(b) the number of revolutions made by the wheel is 800 revolutions.

Explanation:

Given;

mass of the grinding wheel = 1.15 kg

diameter of the wheel = 22.0 cm

number of revolution = 20

time taken to come to rest = 80.0 s

Part (a) the angular acceleration (in rad/s2) of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive.

$\omega _f = \omega_i + \alpha t$

where;

ωf is the final angular speed

ωi is the initial angular speed

α is the angular acceleration

$\omega _f = \omega_i + \alpha t\\\\0 = (2\pi *20) + 80 \alpha\\\\ 80 \alpha = -40\pi \\\\\alpha = \frac{-40\pi}{80} = -1.571 \frac{rad}{s^2}$

Part (b) number of revolutions the wheel made during the time it was coming to rest

$\theta = \omega_i t+\frac{1}{2} \alpha t^2\\\\ \theta = \frac{1}{2}(\frac{1.571}{2\pi } ) (80)^2 = 800, revolutions$