Question

A wheel with radius r rotates freely about its axle with initial angular speed ωi. A point on the wheel's rim moves at linear speed v.

Suddenly, the wheel accelerates so the point moves at linear speed 3v. The final angular speed is ωf.

How does the angular speed of the wheel change?

Answer 1

Answer:

$\dfrac{2\omega_i}{t}$

Explanation:

r = Radius of wheel

v = Initial linear speed

3v = Final linear speed

Initial angular speed is given by

$\omega_i=\dfrac{v}{r}$

Final angular speed after time $t$ is given by

$\omega_f=\dfrac{3v}{r}\\\Rightarrow \omega_f=3\dfrac{v}{r}=3\omega_i$

The angular acceleration of the wheel will give us how the angular speed of the wheel changes.

From the equations of rotational motion we have

$\omega_f=\omega_i+\alpha t\\\Rightarrow 3\omega_i=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{3\omega_i-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{2\omega_i}{t}$

The acceleration wheel is $\dfrac{2\omega_i}{t}$.

Answer 2

Answer:increases by a factor of 3

Explanation: just trust me bro