Question

A wheel of diameter 23.0 cm is constrained to rotate in the xy plane, about the z axis, which passes through its center. A force =( -31.0 + 40.0) N acts at a point on the edge of the wheel that lies exactly on the x axis at a particular instant.What is the magnitude of the torque about the rotation axis at this instant?

Answer 1

The magnitude of the torque on a wheel with a diameter of 23.0 cm when a force acts on its edge in the xy-plane is 4.6 Nm.

Step-by-step explanation:

The question asks for the magnitude of the torque exerted on a wheel when a force acts on its edge in the xy-plane. The diameter of the wheel is given as 23.0 cm, which means its radius is half, i.e., 11.5 cm or 0.115 m. The force vector given is (-31.0 + 40.0) N, but for torque calculation around the z-axis, only the force component perpendicular to the radius vector contributes to torque. Since the force acts on the x-axis, only the y-component of the force is perpendicular. Thus, the torque (τ) is calculated using the formula τ = r * F * sin(θ), where θ is the angle between the force vector and the position vector (radius in this case). Here, the angle is 90° as the force acts along the y-axis while the radius is along the x-axis. The sin(90°) is 1, which makes the calculation straightforward: τ = 0.115 m * 40.0 N * 1 = 4.6 Nm. The direction of the torque would be out of the plane (along the z-axis) according to the right-hand rule. Therefore, the magnitude of the torque about the rotation axis at this instant is 4.6 Nm.

Answer 2

τ=9.2 N.m

Step-by-step explanation:

The component that contributes to torque is the y-component of the force F, so:

τ = r x F = (0.23 m) * (40 N) = 9.2m

Expressed as a vector equation:

τ = [0.23,0] x [ -31, 40] = [0, 0, 9.2] N.m