Question

The wheel of a stationary exercise bicycle at your gym makes one rotation in 0.670 s. Consider two points on this wheel: Point P is 10.0 cm from the rotation axis, and point Q is 20.0 cm from the rotation axis. Find (a) the speed of each point on the spinning wheel and (b) the centripetal acceleration of each point. (c) For points on this spinning wheel, as the distance from the axis increases, does the speed increase or decrease

Answer 1

Answer:

a) For P: $v=0.938\frac{m}{s}$

For Q: $v = 1.876\frac{m}{s}$

b) For P:

$a_{rad}=8.80\frac{m}{s^{2}}$

for Q:

$a_{rad}=17.60\frac{m}{s^{2}}$

c) As the distance from the axis increases then speed increases too.

Explanation:

a) Assuming constant angular acceleration we can find the angular speed of the wheel dividing the angular displacement θ between time of rotation:

$\omega =\frac{\theta}{t}$

One rotation is 360 degrees or 2π radians, so θ=2π

$\omega =\frac{2\pi}{0.670} =9.38\frac{rad}{s}$

Angular acceleration is at every point on the wheel, but speed (tangential speed) is different and depends on the position (R) respect the rotation axis, the equation that relates angular speed and speed is:

$v = \omega R$

for P:

$v = 9.38\frac{rad}{s}*0.1m=0.938\frac{m}{s}$

for Q:

$v = 9.38\frac{rad}{s}*0.2m=1.876\frac{m}{s}$

b) Centripetal acceleration is:

$a_{rad}= \frac{v^2}{R}$

for P:

$a_{rad}= \frac{(0.938)^2}{0.1}=8.80\frac{m}{s^{2}}$

for Q:

$a_{rad}= \frac{(1.876)^2}{0.2}=17.60\frac{m}{s^{2}}$

c) As seen on a) speed and distance from axis is $v = \omega R$ because ω is constant the if R increases then v increases too.