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Chapter 10: Problem 48

A pulley \(12 \mathrm{cm}\) in diameter is free to rotate about a horizontal axle. A \(220-\mathrm{g}\) mass and a \(470-\mathrm{g}\) mass are tied to either end of a massless string, and the string is hung over the pulley. Assuming the string doesn't slip, what torque must be applied to keep the pulley from rotating?

Short Answer

Expert verified
In order to prevent the pulley from rotating, a torque of \(0.146 \, \mathrm{N \cdot m}\) must be applied.

Step by step solution

01

Convert Mass to Weight

Firstly, convert the masses from grams to kilograms and then to weights. Using the formula \( W = m \cdot g \), with \( g \) as \( 9.8 \, \mathrm{m/s^2} \), we get weights of the masses \( W_1 \) and \( W_2 \) as \( 0.22 \, \mathrm{kg} \cdot 9.8 \, \mathrm{m/s^2} = 2.156 \, \mathrm{N} \) and \( 0.47 \, \mathrm{kg} \cdot 9.8 \, \mathrm{m/s^2} = 4.606 \, \mathrm{N} \) respectively.
02

Convert Pulley Diameter to Radius

Next, the diameter of the pulley is given as 12 cm. The radius of the pulley \( r \) is thus half of the diameter: \( \frac{12 \, \mathrm{cm}}{2} = 6 \, \mathrm{cm} = 0.06 \, \mathrm{m} \).
03

Calculate Individual Torques from Each Mass

Now the individual torques from each mass \( \Gamma_1 \) and \( \Gamma_2 \) can be calculated using the formula \( \Gamma = r \cdot F \cdot sin(\u03b8) \). Assuming the angle \( \u03b8 \) is 90 degrees (the tension will be pulling straight down the rope, perpendicular to the radius), and using given forces \( F = W_1, W_2 \) and radius \( r \), we calculate \( \Gamma_1 = 0.06 \, \mathrm{m} \cdot 2.156 \, \mathrm{N} = 0.130 \, \mathrm{N \cdot m} \) and \( \Gamma_2 = 0.06 \, \mathrm{m} \cdot 4.606 \, \mathrm{N} = 0.276 \, \mathrm{N \cdot m} \).
04

Find Equilibrium Torque

Given that the system is not rotating, these two torques are equal in magnitude and opposite in direction. So we can equate them, solve for the torque to be applied to maintain equilibrium, \( \Gamma = \Gamma_2 - \Gamma_1 = 0.276 \, \mathrm{N \cdot m} - 0.130 \, \mathrm{N \cdot m} = 0.146 \, \mathrm{N \cdot m} \).

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Most popular questions from this chapter

Each propeller on a King Air twin-engine airplane consists of three blades, each of mass \(10 \mathrm{kg}\) and length \(125 \mathrm{cm} .\) The blades may be treated approximately as uniform, thin rods, (a) What's the propeller's rotational inertia? (b) If the plane's engine develops a torque of \(2.7 \mathrm{kN} \cdot \mathrm{m}\), how long will it take to spin up the propeller from 1400 rpm to 1900 rpm?

You're skeptical about a new hybrid car that stores energy in a flywheel. The manufacturer claims the flywheel stores 12 MJ of energy and can supply \(40 \mathrm{kW}\) of power for 5 minutes. You dig deeper and find that the flywheel is a \(39-\) cm-diameter ring with mass \(48 \mathrm{kg}\) that rotates at 30,000 rpm. Are the specs correct?

You're an engineer designing kitchen appliances, and you're working on a two- speed food blender, with 3600 rpm and 1800 rpm settings. Specs call for the blender to make no more than 60 revolutions while it's switching from high to low speed. If it takes \(1.4 \mathrm{s}\) to make the transition, does it meet its specs?

A \(110-\mathrm{N} \cdot \mathrm{m}\) torque is needed to start a revolving door rotating. If a child can push with a maximum force of \(90 \mathrm{N},\) how far from the door's rotation axis must she apply this force?

Is it possible to apply a counterclockwise torque to an object that's rotating clockwise? If so, how will the object's motion change? If not, why not?
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