It is desired that the outer edee of a grindine wheel 9.0 cm in radius move at a rate of 6.0 m/s. (a) Determine the angular speed of the wheel. (b) What length of thread could be wound on the rim of the wheel in 3.0s when it is turning at this rate

3 answers

(a) The linear velocity of a point on the edge of the wheel is given by v = rω, where r is the radius and ω is the angular speed. Rearranging this equation, we get ω = v/r.

Substituting the given values, we get:

ω = 6.0 m/s / 0.09 m = 66.7 rad/s

Therefore, the angular speed of the wheel is 66.7 rad/s.

(b) The length of thread wound on the rim of the wheel in 3.0 s is given by L = rθ, where r is the radius and θ is the angular displacement of the wheel in radians, which is equal to ωt.

Substituting the given values, we get:

θ = ωt = 66.7 rad/s × 3.0 s = 200.1 rad

L = 0.09 m × 200.1 rad = 18.01 m

Therefore, the length of thread that could be wound on the rim of the wheel in 3.0 s when it is turning at this rate is 18.01 m.
A force of 200 N acts tangentially on the rim of a wheel 25 cm in radius. (a) Find the torque. (b) Repeat if
the force makes an angle of 40 to a spoke of the wheel.
(a) The torque τ produced by the tangential force F at a distance r from the center of the wheel is given by τ = Fr. Substituting the given values, we get:

τ = 200 N × 0.25 m = 50 Nm

Therefore, the torque produced by the force is 50 Nm.

(b) If the force makes an angle of 40° to a spoke of the wheel, then the lever arm is the perpendicular distance from the line of action of the force to the axis of rotation, which is given by r sin θ, where r is the radius of the wheel and θ is the angle between the force and the spoke.

Substituting the given values, we get:

r sin θ = 0.25 m × sin 40° ≈ 0.16 m

The torque is now given by τ = Fr sin θ. Substituting the given values, we get:

τ = 200 N × 0.16 m ≈ 32 Nm

Therefore, the torque produced by the force at an angle of 40° to a spoke of the wheel is approximately 32 Nm.