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Chapter 17: Problem 113

The uniform disk of mass \(m\) is rotating with an angular velocity of \(\omega_{0}\) when it is placed on the floor. Determine the initial angular acceleration of the disk and the acceleration of its mass center. The coefficient of kinetic friction between the disk and the floor is \(\mu_{k}\).

Short Answer

Expert verified
The initial angular acceleration of the disk is \(2 \mu_{k} \cdot g / r\) and the acceleration of its mass center is \(\mu_{k} \cdot g\).

Step by step solution

01

Identify Knowns and Unknowns

The known variables are the disk's mass \(m\), initial angular velocity \(\omega_{0}\), and the coefficient of kinetic friction \(\mu_{k}\). The unknowns are the initial angular acceleration and the acceleration of the disk's mass center.
02

Express Torque in Terms of Friction

The torque acting on the disk is due to the friction between the disk and the floor. It can be expressed as: \(\tau = f_{friction} \cdot r\), where \(f_{friction}\) is given by \(\mu_{k} \cdot m \cdot g\), \(g\) is the acceleration due to gravity, and \(r\) is the radius of the disk.
03

Use Newton's Second Law for Rotation

Newtons second law for rotation states that \(\tau = I \cdot \alpha\), where \(\tau\) is the torque, \(\alpha\) is the angular acceleration, and \(I\) is the moment of inertia. For a disk, \(I = 0.5 \cdot m \cdot r^2\). Substituting the expression for torque, we get \(\mu_{k} \cdot m \cdot g \cdot r = 0.5 \cdot m \cdot r^2 \cdot \alpha\), which solves to \(\alpha = 2 \mu_{k} \cdot g / r\).
04

Use Newton's Second Law for Linear Motion

Newton's second law for linear motion states that \(f_{net} = m \cdot a_{cm}\), where \(a_{cm}\) is the acceleration of the mass center of the disk. The only horizontal force acting on the disk is the friction force, therefore the acceleration of the disk's mass center (\(a_{cm}\)) equals \(f_{friction} / m = \mu_{k} \cdot g\).

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Most popular questions from this chapter

The \(20-\mathrm{kg}\) roll of paper has a radius of gyration \(k_{A}=120 \mathrm{~mm}\) about an axis passing through point \(A .\) It is pin supported at both ends by two brackets \(A B .\) The roll rests on the floor, for which the coefficient of kinetic friction is \(\mu_{k}=0.2\). If a horizontal force \(F=60 \mathrm{~N}\) is applied to the end of the paper, determine the initial angular acceleration of the roll as the paper unrolls.

Cable is unwound from a spool supported on small rollers at \(A\) and \(B\) by exerting a force \(T=300 \mathrm{~N}\) on the cable. Compute the time needed to unravel \(5 \mathrm{~m}\) of cable from the spool if the spool and cable have a total mass of \(600 \mathrm{~kg}\) and a radius of gyration of \(k_{O}=1.2 \mathrm{~m}\). For the calculation, neglect the mass of the cable being unwound and the mass of the rollers at \(A\) and \(B .\) The rollers turn with no friction.

Determine the moment of inertia of the homogeneous triangular prism with respect to the \(y\) axis. Express the result in terms of the mass \(m\) of the prism. Hint: For integration, use thin plate elements parallel to the \(x-y\) plane and having a thickness \(d z\).

The \(10-\mathrm{kg}\) wheel has a radius of gyration \(k_{A}=200 \mathrm{~mm}\). If the wheel is subjected to a moment \(M=(5 t) \mathrm{N} \cdot \mathrm{m}\), where \(t\) is in seconds, determine its angular velocity when \(t=3 \mathrm{~s}\) starting from rest. Also, compute the reactions which the fixed pin \(A\) exerts on the wheel during the motion.

The sphere is formed by revolving the shaded area around the \(x\) axis. Determine the moment of inertia \(I_{x}\) and express the result in terms of the total mass \(m\) of the sphere. The material has a constant density \(\rho\).
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