Question

A cyclist uses a stationary trainer during the winter to keep in shape. Knowing that she pushes down on her pedal with a velocity of 26 in./s that increases at a rate of 2.2 in./s2, determine the velocity and acceleration (tangential and normal) of point D on the bottom of the rear wheel.

Answer 1

Answer: v = 112in/s

              a = 896.05in/s²

Explanation: Angular velocity is the arc traveled per change in time or:

$\omega=\frac{v}{r}$

For the pedal, it will be:

$\omega_{p}=\frac{v_{p}}{r_{p}}$

$\omega_{p}=\frac{26}{6.5}$

$\omega_{p}=4$ rad/s

Angular velocity of the pedal is equal to the angular velocity of the chain:

$\omega_{p}d_{p}=\omega_{s}d_{s}$

where

$d_{p}$ is diameter of the pedal

$d_{s}$ is diameter of the small circle at the rear wheel

$\omega_{s}=\frac{7}{3.5}.4$

$\omega_{s}$ = 8 rad/s

Velocity at the point D at the bottom othe rear wheel:

$v_{D}=\omega_{s}r_{D}$

$v_{D}=$ 8.14

$v_{D}$ = 112in/s

Velocity at point D is 112 in/s.

Acceleration in circular motion is the vectorial sum of normal and tangential acceleration.

Angular acceleration is also known as centripetal acceleration, due to its direction pointing towards the center. It is "responsible" for changing the object direction. It is calculated as $a_{c}=\omega^{2}.r$

Tangential acceleration is the change in magnitude of the circular motion and is calculated as

$a_{t}=r.\frac{\Delta \omega}{\Delta t}$ or $a_{t}=r.\alpha$

Angular and tangential accelerations are directly proportional.

Tangential Acceleration of the pedal is:

$\alpha_{p}=\frac{a_{t}}{r}$

$\alpha_{p}=\frac{2.2}{6.5}$

$\alpha_{p}$ = 0.34 rad/s²

Tangential acceleration at the pedal is equal to angular acceleration at the rear wheel:

$r_{w}.\alpha_{w}=r_{p}.\alpha_{p}$

$r_{w}$ is radius of the small circle at the rear wheel

$r_{p}$ is radius of the pedal

$\alpha_{w}=\frac{3.5}{1.75}.0.34$

$\alpha_{w}$ = 0.68 rad/s²

At the point D, tangential acceleration will be:

$a_{t,D}=\alpha_{w}.r_{D}$

$a_{t,D}=0.68*14$

$a_{td,D}$ = 9.52 in/s²

Centripetal acceleration at point D is:

$a_{c,D}=\omega_{D}^{2}.r_{D}$

$a_{c,D}=8^{2}.14$

$a_{c,D}$ = 896 in/s²

Net acceleration:

$a_{D}=\sqrt{a_{t,D}^{2}+a_{c,D}^{2}}$

$a_{D}=\sqrt{9.52^{2}+896^{2}}$

$a_{D}$ = 896.05 in/s²

Acceleration of point D at the bottom of rear wheel is 896.05 in/s²