To determine the magnitudes of the velocity and acceleration of peg P at $\theta = 30^\circ$, we need to analyze the motion based on the provided angular velocity equation and the constraints of the system. The system involves a rotating arm $OA$ and a constrained motion of peg $P$ along a lemniscate-shaped slot in $OB$. Let's break it down step-by-step.

Given Information:

• Angular velocity of arm $OA$: $\dot{\theta} = \frac{3t^3}{2} \, \text{rad/s}$
• At time $t = 0$, $\theta = 0^\circ$
• We need to find the magnitudes of velocity and acceleration at $\theta = 30^\circ$.

1. Angular Position of Arm $OA$ (as a function of time)

We know that angular velocity is the derivative of the angular position: $\dot{\theta} = \frac{3t^3}{2}$ To find $\theta(t)$, we integrate the angular velocity with respect to time: $\theta(t) = \int \dot{\theta} \, dt = \int \frac{3t^3}{2} \, dt = \frac{3t^4}{8} + C$ Given that at $t = 0$, $\theta = 0$, we can find the constant of integration $C$: $0 = \frac{3(0)^4}{8} + C \quad \Rightarrow \quad C = 0$ Thus, the angular position of the arm as a function of time is: $\theta(t) = \frac{3t^4}{8}$

2. Find Time Corresponding to $\theta = 30^\circ$

We need to find the time $t$ when $\theta = 30^\circ$: $\theta(t) = 30^\circ = \frac{3t^4}{8}$ To convert degrees to radians, we use: $30^\circ = \frac{\pi}{6} \, \text{rad}$ Now solve for $t$: $\frac{\pi}{6} = \frac{3t^4}{8} \quad \Rightarrow \quad t^4 = \frac{8\pi}{18} = \frac{4\pi}{9}$ $t = \left(\frac{4\pi}{9}\right)^{1/4}$

3. Velocity of Peg P

The velocity of peg $P$ can be found by considering the instantaneous velocity relative to the rotating arm. The tangential velocity $v$ of peg $P$ is given by: $v = r \cdot \dot{\theta}$ where $r$ is the distance of peg $P$ from the center of rotation (point $O$).

To calculate the exact value of $v$, we would need the value of $r$, which is typically a known parameter in the problem. If we assume that $r$ is constant or can be determined from the geometry, we can plug in the value for $\dot{\theta}$ at the specific time $t$.

4. Acceleration of Peg P

The total acceleration of peg $P$ has two components:

1. Tangential Acceleration $a_{\text{tan}} = r \cdot \ddot{\theta}$
2. Centripetal (Radial) Acceleration $a_{\text{rad}} = r \cdot \dot{\theta}^2$

The total acceleration $a$ is the vector sum of these two components. To find the magnitude, we use: $a = \sqrt{a_{\text{tan}}^2 + a_{\text{rad}}^2}$

Where:

• $\ddot{\theta}$ is the angular acceleration, which is the derivative of $\dot{\theta}$: $\ddot{\theta} = \frac{d}{dt} \left( \frac{3t^3}{2} \right) = \frac{9t^2}{2}$

5. Final Steps

To summarize:

1. We find the time corresponding to $\theta = 30^\circ$.
2. Compute the values for $v$ (velocity) and $a$ (acceleration) using the formulas.

Let me know if you'd like further details, clarification, or if you'd like me to work through these final steps numerically.