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Chapter 8: Problem 12

The pull cord of a lawnmower engine is wound around a drum of radius $6.00 \mathrm{cm} .\( While the cord is pulled with a force of \)75 \mathrm{N}$ to start the engine, what magnitude torque does the cord apply to the drum?

Short Answer

Expert verified
Answer: The magnitude of the torque applied to the drum by the pull cord is 4.50 N⋅m.

Step by step solution

01

Identify the given information

We have the following information given in the problem: - Radius of the drum: \(r = 6.00\,\text{cm}\) - Force applied to the pull cord: \(F = 75\,\text{N}\)
02

Convert the radius to meters

To calculate the torque, we need the radius in meters. To convert the radius from centimeters to meters, we can use the following conversion: \(1 \, \mathrm{m} = 100\, \mathrm{cm}\) Therefore, the radius in meters is: \(r = 6.00\, \mathrm{cm} \times \frac{1 \, \mathrm{m}}{100 \, \mathrm{cm}} = 0.0600\,\mathrm{m}\)
03

Calculate the torque

Now, we can use the formula for torque, which is: \(\tau = r \times F\) Plugging in the values for the radius and the force, we get: \(\tau = (0.0600 \, \mathrm{m}) \times (75 \, \mathrm{N})\) Calculating the torque: \(\tau = 4.50\,\mathrm{N} \cdot \mathrm{m}\)
04

State the result

The magnitude of the torque applied to the drum by the pull cord is \(4.50\,\mathrm{N} \cdot \mathrm{m}\).

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Most popular questions from this chapter

A merry-go-round (radius \(R\), rotational inertia \(I_{\mathrm{i}}\) ) spins with negligible friction. Its initial angular velocity is \(\omega_{i}\) A child (mass \(m\) ) on the merry-go-round moves from the center out to the rim. (a) Calculate the angular velocity after the child moves out to the rim. (b) Calculate the rotational kinetic energy and angular momentum of the system (merry-go-round + child) before and after.
A bowling ball made for a child has half the radius of an adult bowling ball. They are made of the same material (and therefore have the same mass per unit volume). By what factor is the (a) mass and (b) rotational inertia of the child's ball reduced compared with the adult ball?
Derive the rotational form of Newton's second law as follows. Consider a rigid object that consists of a large number \(N\) of particles. Let \(F_{i}, m_{i},\) and \(r_{i}\) represent the tangential component of the net force acting on the ith particle, the mass of that particle, and the particle's distance from the axis of rotation, respectively. (a) Use Newton's second law to find \(a_{i}\), the particle's tangential acceleration. (b) Find the torque acting on this particle. (c) Replace \(a_{i}\) with an equivalent expression in terms of the angular acceleration \(\alpha\) (d) Sum the torques due to all the particles and show that $$\sum_{i=1}^{N} \tau_{i}=I \alpha$$
Verify that the units of the rotational form of Newton's second law [Eq. (8-9)] are consistent. In other words, show that the product of a rotational inertia expressed in \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular acceleration expressed in \(\mathrm{rad} / \mathrm{s}^{2}\) is a torque expressed in \(\mathrm{N} \cdot \mathrm{m}\).
A 68 -kg woman stands straight with both feet flat on the floor. Her center of gravity is a horizontal distance of \(3.0 \mathrm{cm}\) in front of a line that connects her two ankle joints. The Achilles tendon attaches the calf muscle to the foot a distance of \(4.4 \mathrm{cm}\) behind the ankle joint. If the Achilles tendon is inclined at an angle of \(81^{\circ}\) with respect to the horizontal, find the force that each calf muscle needs to exert while she is standing. [Hint: Consider the equilibrium of the part of the body above the ankle joint.]
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