Question

1. lever arm you open a door by pushing at a right angle to the door. your friend pushes at the same place, but at an angle of 55° from the perpendicular. if both you and your friend exert the same torque on the door, how do the forces you and your friend applied compare?33. the solid wheel shown in figure 15 has a mass of 5.2 kg and a diameter of 0.55 m. it is at rest, and you need it to rotate at 12 rev/s in 35 s. a. what torque do you need to apply to the wheel? b. if a nylon strap is wrapped around the outside of the wheel, how much force do you need to exert on the strap?figure 15 image of a solid wheel with a force f applied horizontally to the left at the top, and a torque τ symbol indicating rotation

Explanation:

Problem 32: Lever Arm

Step 1: Recall Torque Formula

Torque \(\tau\) is given by \(\tau = rF\sin\theta\), where \(r\) is the distance from the pivot, \(F\) is the force, and \(\theta\) is the angle between \(r\) and \(F\).
For you, \(\theta_1 = 90^\circ\) (right angle), so \(\sin\theta_1 = \sin 90^\circ = 1\). Your torque: \(\tau_1 = rF_1(1) = rF_1\).
For your friend, \(\theta_2 = 55^\circ\), so \(\sin\theta_2 = \sin 55^\circ\). Friend's torque: \(\tau_2 = rF_2\sin 55^\circ\).

Step 2: Set Torques Equal

Given \(\tau_1 = \tau_2\), so \(rF_1 = rF_2\sin 55^\circ\). Cancel \(r\) (same place, so \(r\) is equal): \(F_1 = F_2\sin 55^\circ\). Rearrange to find \(F_2\) in terms of \(F_1\): \(F_2 = \frac{F_1}{\sin 55^\circ}\). Calculate \(\sin 55^\circ \approx 0.8192\), so \(F_2 \approx \frac{F_1}{0.8192} \approx 1.22F_1\).

Step 1: Find Moment of Inertia for Solid Disk

Moment of inertia \(I\) for a solid disk (wheel) is \(I = \frac{1}{2}MR^2\), where \(M = 5.2\,\text{kg}\) and radius \(R = \frac{d}{2} = \frac{0.55\,\text{m}}{2} = 0.275\,\text{m}\).
Calculate \(I\): \(I = \frac{1}{2}(5.2\,\text{kg})(0.275\,\text{m})^2\).
First, \((0.275)^2 = 0.075625\), then \(I = 0.5 \times 5.2 \times 0.075625 = 2.6 \times 0.075625 = 0.196625\,\text{kg·m}^2\).

Step 2: Find Angular Acceleration (\(\alpha\))

Initial angular velocity \(\omega_0 = 0\) (at rest), final angular velocity \(\omega = 12\,\text{rev/s}\). Convert to rad/s: \(\omega = 12 \times 2\pi = 24\pi\,\text{rad/s}\) (or \(75.398\,\text{rad/s}\)). Time \(t = 35\,\text{s}\).
Angular acceleration \(\alpha = \frac{\omega - \omega_0}{t} = \frac{24\pi}{35}\,\text{rad/s}^2 \approx \frac{75.398}{35} \approx 2.154\,\text{rad/s}^2\).

Step 3: Calculate Torque (\(\tau\))

Torque \(\tau = I\alpha\). Substitute \(I = 0.196625\,\text{kg·m}^2\) and \(\alpha \approx 2.154\,\text{rad/s}^2\): \(\tau = 0.196625 \times 2.154 \approx 0.424\,\text{N·m}\) (more precisely, using \(\omega = 24\pi\)): \(\alpha = \frac{24\pi}{35}\), so \(\tau = 0.196625 \times \frac{24\pi}{35} \approx 0.196625 \times 2.1539 \approx 0.423\,\text{N·m}\)).

Step 1: Recall Torque - Force Relationship

Torque \(\tau = rF\sin\theta\). Here, the force is applied tangentially (since the strap is wrapped around the wheel), so \(\theta = 90^\circ\), \(\sin\theta = 1\). Radius \(r = R = 0.275\,\text{m}\) (distance from center to edge, where force is applied).

Step 2: Solve for Force

From \(\tau = rF\), rearrange: \(F = \frac{\tau}{r}\). We found \(\tau \approx 0.424\,\text{N·m}\) (from part a), \(r = 0.275\,\text{m}\). So \(F = \frac{0.424}{0.275} \approx 1.54\,\text{N}\) (using \(\tau \approx 0.423\,\text{N·m}\), \(F = \frac{0.423}{0.275} \approx 1.54\,\text{N}\)).

Answer:

Your friend's force (\(F_2\)) is approximately \(1.22\) times your force (\(F_1\)) (or \(F_{\text{friend}} \approx 1.22F_{\text{you}}\)).

Problem 33a: Torque on Solid Wheel