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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 centrifuge has a rotational inertia of $6.5 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}$ How much energy must be supplied to bring it from rest to 420 rad/s \((4000 \text { rpm }) ?\)
The string in a yo-yo is wound around an axle of radius \(0.500 \mathrm{cm} .\) The yo-yo has both rotational and translational motion, like a rolling object, and has mass \(0.200 \mathrm{kg}\) and outer radius \(2.00 \mathrm{cm} .\) Starting from rest, it rotates and falls a distance of \(1.00 \mathrm{m}\) (the length of the string). Assume for simplicity that the yo-yo is a uniform circular disk and that the string is thin compared to the radius of the axle. (a) What is the speed of the yo-yo when it reaches the distance of $1.00 \mathrm{m} ?$ (b) How long does it take to fall? [Hint: The transitional and rotational kinetic energies are related, but the yo-yo is not rolling on its outer radius.]
What is the rotational inertia of a solid iron disk of mass \(49 \mathrm{kg},\) with a thickness of \(5.00 \mathrm{cm}\) and radius of \(20.0 \mathrm{cm}\) about an axis through its center and perpendicular to it?
A uniform diving board, of length \(5.0 \mathrm{m}\) and mass \(55 \mathrm{kg}\) is supported at two points; one support is located \(3.4 \mathrm{m}\) from the end of the board and the second is at \(4.6 \mathrm{m}\) from the end (see Fig. 8.19 ). What are the forces acting on the board due to the two supports when a diver of mass \(65 \mathrm{kg}\) stands at the end of the board over the water? Assume that these forces are vertical. (W) tutorial: plank) [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.]
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.
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