Question 12.SP.9: A 0.5-kg collar is attached to a spring and slides without f......
A 0.5-kg collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 150 mm and a constant k = 200 N/m. Knowing that the collar has a speed of 3 m/s as it passes through point B, determine the tangential acceleration of the collar and the force of the rod on the collar at this instant.
STRATEGY: This problem deals with forces and accelerations, so you need to use Newton’s second law. The collar moves along a curved path, so you should use normal and tangential coordinates.
MODELING: Choose the collar as your system and assume you can treat it as a particle. Draw the free-body and kinetic diagrams as shown in Fig. 1. The spring force acts in the direction of the spring, and the force is drawn assuming that the spring is stretched and not compressed. Check this using geometry.
\tan\alpha={\frac{125\;{\mathrm{mm}}}{300\,{\mathrm{mm}}}}=0.4167\to\alpha=22.62^{\circ}
L_{B D}=\sqrt{(300\;\mathrm{mm})^{2}+(125\;\mathrm{mm})^{2}}=325\;\mathrm{mm}
Thus, when the collar is at B, the spring is extended as
x=L_{B D}-L_{0}=325\;\mathrm{mm}-150\;\mathrm{mm}=175\;\mathrm{mm}.ANALYSIS: You can obtain scalar equations by applying Newton’s second law in the normal and tangential directions. Hence,
+↑\Sigma F_{n}=m a_{n}\qquad\qquad k x\sin\alpha+N-m g=m a_{n}=m{\frac{\nu^{2}}{\rho}} (1)
\underleftarrow{+}\Sigma F_{t}=m a_{t}\qquad\qquad F x\cos\alpha=m a_{t} (2)
You now have two equations, (1) and (2), and two unknowns, a_{t} and N. You can solve for these by hand or using your calculator/computer. You can solve for the normal force in Eq. (1) as
N=m g+m{\frac{\nu^{2}}{\rho}}-k x\sin\alpha
Substituting values gives
N = (0.5 kg)(9.81 m/s²) + (0.5 kg)\frac{(3\ m/s)^{2}}{0.125\ m}
− (200 N/m)(0.175 m) sin (22.62°)
N = 27.4 N ◂
a_{t}={\frac{F x\cos\alpha}{m}}={\frac{(200\ \mathrm{N/m})(0.175)\cos\left(22.62^{\circ}\right)}{0.5\ \mathrm{kg}}}a_{t}= 64.6 m/s² ◂
REFLECT and THINK: How would this problem have changed if you had been told friction was acting between the rod and the collar? You would have had one additional term in your free-body diagram, \mu_{k}N, in the direction opposite to the velocity. Thus, you would need to be told the direction the collar was moving as well as the coefficient of kinetic friction.
