Chapter 10, Problem 98 A yo-yo-shaped device mounted on a horizontal frictionles

Question

Chapter 10, Problem 98 A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a 28 kg box as shown in the figure. The outer radius R of the device is 0.46 m, and the radius r of the hub is 0.14 m. When a p of magnitude 120 N is applied to a rope wrapped around the outside of the device, the box, which is suspended from a rope wrapped constant horizontal force around the hub, has an upward acceleration of magnitude 0.74 m/s2. What is the rotational inertia of the device about its axis of rotation? Rigid mount Hub Yoyo shaped device . Rope Number Units

Explanation / Answer

The device rotates in counterclockwise direction in order for the box to move up.

Take counterclockwise direction as positive.

Let,

alpha = angular acceleration of the device,

a = linear acceleration of a point on the rim of the hub

a = acceleration of the box = 0.74 m/s2

alpha = a/r = 0.74 / 0.14

Or alpha = 5.28 rad/s2 -------------------------------------(1)

Let T = tension in the rope

Forces on the box are

1. Tension T upward

2. Weigth mg downward

Net upward force = T - mg

Upward acceleration of box = a

Therefore, T - mg = ma

Or T = mg + ma = m(g + a)

Or T = 28*(9.8 + 0.74)

T = 295.12 N---------------------------(2)

Calculate torques on the device around the axis of rotation taking counterclockwise torque as +ve and clockwise as -ve.

Torque by applied force F = F*R = 120 * 0.46 = 55.2 Nm

Torque by T = -T r = -295.12 * 0.14 [using value of T from equation(2)]

= - 41.32 Nm

Net torque on the device = 55.2 - 41.32 Nm = 13.9 Nm-------(3)

Let rotational inertial of the device = I

I = net torque/alpha

Using equations (1) and (3) in the above,

I = 13.9/5.28 kg-m2

I = 2.63 kg-m2

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