Chapter 12.1, Problem 12.65P

A small 250-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at a constant rate of 7.5 rad/s. Knowing that the coefficients of friction are μ_s = 0.25 and μ_k = 0.20, indicate whether the collar will slide on the rod if it is released in the position corresponding to (a) θ = 75°, (b) θ = 40°. Also, determine the magnitude and direction of the friction force exerted on the collar immediately after release.

Fig. P12.64 and P12.65

To determine

Indicate whether the collar will slide on the rod if it is released in the position corresponding to angle $\theta$ is $75°$.

Find the magnitude and direction of the friction force exerted on the collar immediately after release.

Answer to Problem 12.65P

The collar does not slide on the rod if it is released in the position corresponding to angle $\theta$ is $75°$.

The magnitude and direction of the friction force exerted on the collar immediately after release is 0.611 N.

Explanation of Solution

Given information:

The mass of collar (m) is 250 g.

The speed of rotation of semi-circular rod $\left(\omega \right)$ is 7.5 rad/s.

The radius (r) of semicircular rod is 500 mm.

The coefficient of static friction $\left({\mu }_s\right)$ between collar and rod is 0.25.

The coefficient of kinetic friction $\left({\mu }_k\right)$ between collar and rod is 0.20.

Calculation:

The collar will not slide. Therefore the collar moves at constant speed.

Find the equation of radius of circle for constant rotation.

$\rho =r\sin \theta$

Find the equation of speed of semicircular rod for constant rotation:

$v=\rho \omega$

Find the equation of normal acceleration $\left(a_n\right)$.

$a_n=\frac{v^2}{\rho }$

Substitute $\rho \omega$ for v.

$a_n=\frac{{\left(\rho \omega \right)}^2}{\rho }$

Substitute $r\sin \theta$ for $\rho$.

$\begin{array}{c}{a}_n=\frac{{\left(r\sin \theta \right)}^2{\omega }^2}{r\sin \theta }\\ =\left(r\sin \theta \right){\omega }^2\end{array}$

Sketch the free body diagram and kinetic diagram of the collar C as shown in Figure (1).

Refer Figure (1):

Find the normal force (N) on the collar.

Apply Newton’s law of equation along x-axis.

$\begin{array}{l}\Sigma F_n=ma\\ N-mg\cos \theta =m{a}_n\sin \theta \end{array}$

Substitute $\left(r\sin \theta \right){\omega }^2$ for $a_n$.

$\begin{array}{l}N-mg\cos \theta =m\left[\left(r\sin \theta \right){\omega }^2\right]\sin \theta \\ N=mg\cos \theta +mr{\omega }^2{\sin }^2\theta \\ N=m\left[g\cos \theta +r{\omega }^2{\sin }^2\theta \right]\end{array}$ (1)

Substitute 250 g for m, $9.81{\text{m/s}}^2$ for g, 500 mm for r, 7.5 rad/s for $\omega$, and $75°$ for $\theta$.

$\begin{array}{c}N=\left(250\text{g}\times \frac{1\text{kg}}{1,000\text{g}}\right)\left[9.81\cos \left(75°\right)+\left(500\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right){\left(7.5\right)}^2{\sin }^2\left(75°\right)\right]\\ =7.1950\text{N}\end{array}$

Find the frictional force (F) using Newton’s law of equation:

Apply Newton’s law of equation along y-axis.

$\begin{array}{l}\Sigma F_y=ma\\ F-mg\sin \theta =m\left(-a_n\cos \theta \right)\end{array}$

Substitute $\left(r\sin \theta \right){\omega }^2$ for $a_n$.

$\begin{array}{l}F-mg\sin \theta =-m\left(r\sin \theta \right){\omega }^2\cos \theta \\ F=mg\sin \theta -m\left(r\sin \theta \right){\omega }^2\cos \theta \\ F=m\left(g-r{\omega }^2\cos \theta \right)\sin \theta \end{array}$ (2)

Substitute 250 g for m, $9.81m/s^2$ for g, 500 mm for r, 7.5 rad/s for $\omega$, and $75°$ for $\theta$.

$\begin{array}{c}F=\left(250\text{g}\times \frac{1\text{kg}}{1,000\text{g}}\right)\left[9.81-\left(500\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right){\left(7.5\right)}^2\cos \left(75°\right)\right]\sin \left(75°\right)\\ =0.611\text{N}\end{array}$

Find the frictional force using general equation:

$F={\mu }_{s}N$

Substitute 0.25 for ${\mu }_s$ and 7.1950 N for N.

$\begin{array}{c}F=\left(0.25\right)\left(7.1950\right)\\ =1.7987\text{N}\end{array}$

The frictional force $\left(\left|F\right|=0.611\right)$ is less than the force $\left({\mu }_{s}N=1.7987\text{N}\right)$. Therefore, the collar does not slide.

Thus, the magnitude and direction of the friction force exerted on the collar immediately after release is 0.611 N.

To determine

Indicate whether the collar will slide on the rod if it is released in the position corresponding to angle $\theta$ is $40°$.

Find the magnitude and direction of the friction force exerted on the collar immediately after release.

Answer to Problem 12.65P

The collar does not slide on the rod if it is released in the position corresponding to angle $\theta$ is $40°$.

The magnitude and direction of the friction force exerted on the collar immediately after release is 0.957 N.

Explanation of Solution

Calculation:

Find the normal force (N) on the collar using Equation (1):

$N=m\left[g\cos \theta +r{\omega }^2{\sin }^2\theta \right]$ (1)

Substitute 250 g for m, $9.81{\text{m/s}}^2$ for g, 500 mm for r, 7.5 rad/s for $\omega$, and $75°$ for $\theta$.

$\begin{array}{c}N=\left(250\text{g}\times \frac{1\text{kg}}{1,000\text{g}}\right)\left[9.81\cos \left(40°\right)+\left(500\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right){\left(7.5\right)}^2{\sin }^2\left(40°\right)\right]\\ =4.7839\text{N}\end{array}$

Find the frictional force (F) using Equation (2):

$F=m\left(g-r{\omega }^2\cos \theta \right)\sin \theta$

Substitute 250 g for m, $9.81m/s^2$ for g, 500 mm for r, 7.5 rad/s for $\omega$, and $40°$ for $\theta$.

$\begin{array}{c}F=\left(250\text{g}\times \frac{1\text{kg}}{1,000\text{g}}\right)\left[9.81-\left(500\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right){\left(7.5\right)}^2\cos \left(40°\right)\right]\sin \left(40°\right)\\ =-1.8858\text{N}\end{array}$

Find the frictional force using general equation:

$F={\mu }_{s}N$

Substitute 0.25 for ${\mu }_s$ and 4.7839 N for N.

$\begin{array}{c}F=\left(0.25\right)\left(4.7839\right)\\ =1.1960\text{N}\end{array}$

The frictional force $\left(\left|F\right|=1.8858\right)$ is greater than the force $\left({\mu }_{s}N=1.1960\text{N}\right)$. Therefore, the collar will slide.

Sketch the free body diagram and kinetic diagram of the collar C which is sliding as shown in Figure (2).

Refer Figure 2:

Find the normal force (N) on the collar.

Apply Newton’s law of equation along x-axis.

$\begin{array}{l}\Sigma F_n=ma\\ N-mg\cos \theta =m{a}_n\sin \theta \end{array}$

Substitute $\left(r\sin \theta \right){\omega }^2$ for $a_n$.

$\begin{array}{l}N-mg\cos \theta =m\left[\left(r\sin \theta \right){\omega }^2\right]\sin \theta \\ N=mg\cos \theta +mr{\omega }^2{\sin }^2\theta \\ N=m\left[g\cos \theta +r{\omega }^2{\sin }^2\theta \right]\end{array}$ (1)

Substitute 250 g for m, $9.81{\text{m/s}}^2$ for g, 500 mm for r, 7.5 rad/s for $\omega$, and $75°$ for $\theta$.

$\begin{array}{c}N=\left(250\text{g}\times \frac{1\text{kg}}{1,000\text{g}}\right)\left[9.81\cos \left(40°\right)+\left(500\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right){\left(7.5\right)}^2{\sin }^2\left(40°\right)\right]\\ =4.7839\text{N}\end{array}$

Find the magnitude and direction of the friction force $\left(\left|F\right|\right)$ exerted on the collar immediately after release using the equation:

$\left|F\right|={\mu }_{k}N$

Substitute 0.20 for ${\mu }_k$ and 4.7839 N for N.

$\begin{array}{c}\left|F\right|=\left(0.20\right)\left(4.7839\right)\\ =0.957\text{N}\end{array}$

Thus, the magnitude and direction of the friction force exerted on the collar immediately after release is 0.957 N.