Question

The drive chain in a bicycle is applying a torque of 0.850 N ∙ m to the wheel of the bicycle. You can treat the wheel as a thin uniform hoop (or ring) with a mass of 0.750 kg and a radius of 33.0 cm. What is the angular acceleration of the wheel?A. 10.4 rad/s2B. 3.43 rad/s2C. 5.20 rad/s2D. 1.06 rad/s2E. 20.8 rad/s2

Answer 1

A) α = 10.4 rad/s²

Step-by-step explanation:

Newton's second law:

F = ma has the equivalent for rotation:

τ = I * α Formula (1)

where:

τ : It is the moment applied to the body. (Nxm)

I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Data

τ = 0.850 N xm : moment or torque applied to the bicycle wheel

m : 0.750 kg : mass of the wheel

R = 33.0 cm= 0.33 m : radius of the wheel

Moment applied to the wheel (I)

The moment of inertia of a thin uniform hoop (or ring) is defined as follows:

I = m* R²

I = 0.750 kg*(0.33 m)²

I = 0.081675 kg*m²

Angular acceleration of the wheel

We replace data in the formula (1):

τ = I * α

0.850 = 0.081675 * α

α= 0.850 / 0.081675

α = 10.4 rad/s²