Question

47) The moment of inertia of a 0.98-kg bicycle wheel rotating about its center is 0.13 kg · m2. What is the radius of this wheel, assuming the weight of the spokes can be ignored?

Answer 1

The moment of inertia of a bicycle wheel rotating about its center can be calculated using the formula I = 0.5 * m * r^2. Rearranging the formula allows us to solve for the radius, which is found to be 0.25 m for the given values.

Step-by-step explanation:

The formula for the moment of inertia of a wheel rotating about its center is given by:

I = 0.5 * m * r²

Where I is the moment of inertia, m is the mass of the wheel, and r is the radius of the wheel.

Given that the moment of inertia is 0.13 kg · m² and the mass of the wheel is 0.98 kg, we can rearrange the formula to solve for the radius:

r = √(2 * I / m) = √(2 * 0.13 / 0.98) = 0.25 m

Therefore, the radius of the bicycle wheel is 0.25 m.

Answer 2

0.36 m

Step-by-step explanation:

The moment of inertial of a wheel can be calculated as

`I=mr^2`

Where r is the radius of the wheel and m is its mass.

Solving the equation for r, we get

`r=\sqrt[]{(I)/(m)}`

So, replacing I by 0.13 kg m2 and m by 0.98 kg, we get

Therefore, the radius of the wheel is 0.36 m