Question

To start her lawn mower, Julie pulls on a cord that is wrapped around a pulley. The pulley has a moment of inertia about its central axis of = 0.550kg..m22 and a radius of 5.00 cm.There is an equivalent frictional torque impeding her pull of τfτf = 0.430 m..N. a)To accelerate the pulley at αα = 4.55 rad/s22, how much torque does Julie need to apply to the pulley? b) How much tension must the rope exert?

Answer 1

To accelerate the pulley at α = 4.55 rad/s², Julie needs to apply a torque of 2.5025 N·m to the pulley. The tension in the rope must be 2.943 N.

Step-by-step explanation:

To solve for the torque needed to accelerate the pulley at a given angular acceleration, we can use the formula:

τ = Iα

where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

Using the given information, we can calculate the torque as follows:

τ = (0.550 kg·m²)(4.55 rad/s²) = 2.5025 N·m

Therefore, Julie needs to apply a torque of 2.5025 N·m to the pulley to accelerate it at a rate of 4.55 rad/s².

To find the tension in the rope, we can use the formula:

T = Iα + τf

where T is the tension, I is the moment of inertia, α is the angular acceleration, and τf is the frictional torque.

Substituting the given values:

T = (0.550 kg·m²)(4.55 rad/s²) + 0.430 N·m = 2.943 N

Therefore, the rope must exert a tension of 2.943 N.