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Chapter 17: Problem 114

The uniform disk of mass \(m\) is rotating with an angular velocity of \(\omega_{0}\) when it is placed on the floor. Determine the time before it starts to roll without slipping. What is the angular velocity of the disk at this instant? The coefficient of kinetic friction between the disk and the floor is \(\mu_{k}\).

Short Answer

Expert verified
The disk starts rolling without slipping after a time \(t = \omega_{0} r / (3 \mu_k g)\). The angular velocity of the disk at this instant is \(\omega = \omega_{0} / 2\).

Step by step solution

01

Determine the Torque

The torque (\(\tau\)) about the point of contact is the product of the frictional force and the radius of the disk, \(f r = \mu_k m g r\). The angular acceleration (\(\alpha\)) can also be obtained from the torque, \(\alpha = \tau / I\), where \(I\) is the moment of inertia. For a disk, \(I = 0.5 m r^2\). Substituting these values gives \(\alpha = 2 \mu_k g / r\).
02

Determine the Time when Disk Starts Rolling without Slipping

The disk will start rolling without slipping when the rate of decrease of angular velocity equals the rate of increase of linear velocity at its center. In other words, when \(\omega_{0} - \alpha t = r \alpha t\). Solving for \(t\) yields \(t = \omega_{0} / (3 \alpha)\). Substitute \(\alpha\) from the first step to get \(t = \omega_{0} r / (3 \mu_k g)\).
03

Determine the Angular Velocity at this Instant

Substitute \(t\) from step 2 into the equation we used in the beginning to determine when rolling without slipping condition is met, \(\omega = \omega_{0} - \alpha t\). Substituting values will yield \(\omega = \omega_{0} / 2\).

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Most popular questions from this chapter

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