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Chapter 10: Problem 56

A motor is connected to a solid cylindrical drum with diameter \(1.2 \mathrm{m}\) and mass \(51 \mathrm{kg} .\) A massless rope is attached to the drum and tied at the other end to a 38 -kg weight, so the rope will wind onto the drum as it turns. What torque must the motor apply if the weight is to be lifted with acceleration \(1.1 \mathrm{m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
The torque that the motor must apply to lift the weight with the given acceleration is approximately 30.6 Nm.

Step by step solution

01

Calculate the moment of inertia of the cylindrical drum

The moment of inertia (I) of a solid cylinder rotating about its central axis can be calculated using the formula \(I = 0.5m_rR^2\), where \(m_r\) is the mass of the cylinder and \(R\) is the radius. So, the moment of inertia of the cylindrical drum is \(I = 0.5 * 51\,kg * (0.6\,m)^2 = 9.18\,kg\,m^2\).
02

Calculate the total force required to lift the mass

The total force required to lift the mass can be found by using the equation \(F = m(a + g)\), where \(m\) is the mass of the weight, \(a\) is the acceleration, and \(g\) is the acceleration due to gravity which is approximately \(9.8\,m/s^2\). Therefore, the total force required to lift the mass is \(F = 38\,kg * (1.1\,m/s^{2} + 9.8\,m/s^{2}) = 415.8\,N\).
03

Calculate the total torque

The total torque (\(T\)) required by the motor can be calculated using the equation \(T = I\alpha +RF\), where \(R\) is the radius of the cylindrical drum and \(F\) is the total force required to lift the weight. \(\alpha = \frac{a}{R}\), is the angular acceleration, so substituting \(a\) gives \(T = 9.18\,kg\,m^2 * \frac{1.1\,m/s^{2}}{0.6\,m} + 0.6\,m * 415.8\,N = 30.6 \,Nm\)

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Most popular questions from this chapter

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