Question 12.SP.10: A block B with a mass m can slide freely on a frictionless a......

STRATEGY: You want to find a force, so use Newton’s second law. The radial distance r of the mass is changing, as is the angular displacement θ, so use radial and transverse coordinates.

MODELING: Choose block B as your system and assume you can model it as a particle. Because all other forces are perpendicular to the plane of the figure, the only force shown acting on B is the force F perpendicular to OA. Draw free-body and kinetic diagrams for block B as shown in Fig. 1.

ANALYSIS:

Equations of Motion. You can obtain scalar equations by applying Newton’s second law in the radial and transverse directions. Hence,

+\Sigma F_{r}=m a_{r}\colon \qquad\qquad 0=m({\ddot{r}}-r{\dot{\theta}}^{2})             (1)

+\Sigma F_{\theta}=m a_{\theta}\colon\qquad\qquad\qquad F=m(r{\ddot{\theta}}+2{\dot{r}}{\dot{\theta}})             (2)

a. Component v_{r} of Velocity. Because v_{r} = \dot{r}, you have

{\ddot{r}}={\dot{v}}_{r}={\frac{dv_{r}}{d t}}={\frac{dv_{r}}{d r}}{\frac{d r}{d t}}=v_{r}{\frac{dv_{r}}{d r}}

After using Eq. (1) to obtain \ddot{r} = r \dot{\theta^{2}} and recalling that \dot{θ} = \dot{θ}_{0}, you can separate the variables to obtain

\nu_{r}d\nu_{r}={\dot{\theta}}_{0}^{2}r d r

Multiply by 2 and integrate from 0 to v_{r}~and~from~r_{0} to r. The result is

\nu_{r}^{2}=\dot{\theta^{2}}_{0}(r^{2}-r_{0}^{2})\qquad\qquad v_{r}=\dot{\theta}_{0}(r^{2}-r_{0}^{2})^{1/2}  ◂

b. Horizontal Force F. Set \dot{\theta}=\dot{\theta}_{0},\ddot{\theta}=0\,\mathrm{and}\,\dot{r}=\nu_{r} in Eq. (2). Then, substitute for v_r the expression obtained in part a. The result is

F=2m\,\dot{\theta}_{0}(r^{2}-r_{0}^{2})^{1/2}\dot{\theta}_{0}\ \;\;\;\;\;F=2m\,\dot{\theta^{2}}_{0}(r^{2}-r_{0}^{2})^{1/2}  ◂

REFLECT and THINK: Introducing radial and transverse components of force and acceleration also involves using components of velocity in the computation. This is still much simpler and more direct than trying to use other coordinate systems. Even though the radial acceleration is zero, the block accelerates relative to the rod with acceleration \ddot{r}.