Problem 64 - Physics for Scientists and Engineers with Modern Physics

Question

(III) An object of mass \(m\) is constrained to move in a circle of radius \(r\). Its tangential acceleration as a function of time is given by \(a_{\tan }=b+c t^{2},\) where \(b\) and \(c\) are constants. If \(v=v_{0}\) at \(t=0,\) determine the tangential and radial components of the force, \(F_{\text {tan }}\) and \(F_{\mathrm{R}}\), acting on the object at any time \(t>0\).

Step-by-Step Solution

Answer

Tangential Force: \( F_{\tan} = m(b + ct^2) \); Radial Force: \( F_{R} = m\frac{(v_0 + bt + \frac{ct^3}{3})^2}{r} \).

Step 1: Identify Known Quantities and Equations

We are given the tangential acceleration, \(a_{\tan} = b + ct^2\). We also know \(v = v_0\) at \(t = 0\), and we seek to find the tangential and radial components of the force. The tangential force can be found using \(F_{\tan} = ma_{\tan}\).

Step 2: Find Tangential Velocity as a Function of Time

The tangential acceleration is \(a_{\tan} = \frac{dv}{dt}\), so integrate to find the velocity:\[\int dv = \int (b + ct^2) \, dt.\]This results in \(v = v_0 + bt + \frac{ct^3}{3}\) by applying the initial condition \(v = v_0\) at \(t = 0\).

Step 3: Calculate Tangential Force

Use the expression for tangential acceleration in the force equation:\[F_{\tan} = ma_{\tan} = m(b + ct^2).\]

Step 4: Find Radial Acceleration and Force

The radial acceleration is given by \(a_R = \frac{v^2}{r}\). Substitute the expression for tangential velocity we found:\[a_R = \frac{(v_0 + bt + \frac{ct^3}{3})^2}{r}\].Then the radial force is:\[F_{R} = ma_R = m\frac{(v_0 + bt + \frac{ct^3}{3})^2}{r}\].

Key Concepts

Tangential Acceleration

Tangential acceleration is the rate of change of tangential velocity with respect to time. In circular motion, it's crucial because it dictates how quickly the speed along the path changes, not the direction. For the given problem, the tangential acceleration is stated as a function of time: \[ a_{\text{tan}} = b + ct^2 \] where \(b\) and \(c\) are constants. This expression means acceleration can change over time due to the \(ct^2\) term. This scenario indicates a non-linear acceleration path, which is common in more complex circular motion problems.

• At \(t=0\), the acceleration equals \(b\).
• As time increases, \(ct^2\) causes the acceleration to grow, showing an increasing tangential effect.

Understanding how tangential acceleration works helps predict how an object accelerates along the circle's path. This insight is necessary to calculate the tangential force exerted on the object as time passes, which is given by:\[ F_{\text{tan}} = ma_{\text{tan}} = m(b+ct^2) \] This equation highlights that the tangential force varies directly with mass and the rate of tangential acceleration.

Radial Force

Radial force, also known as centripetal force, is vital for understanding the forces that act perpendicular to the motion in circular paths. It's responsible for keeping the object moving in a circle by constantly changing the object's direction to point towards the center of the circle. In our problem, the radial acceleration is defined as:\[ a_R = \frac{v^2}{r} \] where \(v\) is tangential velocity and \(r\) is the circle's radius. Radial acceleration is not constant because it's dependent on the square of velocity, which changes as a function of time due to tangential acceleration. After finding the tangential velocity function:\[ v = v_0 + bt + \frac{ct^3}{3} \] we substitute it into the radial acceleration equation:\[ a_R = \frac{(v_0 + bt + \frac{ct^3}{3})^2}{r} \] We then compute the radial force:\[ F_{R} = ma_R = m\frac{(v_0 + bt + \frac{ct^3}{3})^2}{r} \] The radial force changes over time since both \(v\) and \(a_R\) depend on time, showing a dynamic relation in circular motion. Understanding this concept helps to appreciate how continuously changing forces result in uniform circular motion.

Integration for Velocity

In the context of acceleration and velocity, integration plays an essential role in determining the velocity function when given the acceleration function. For this exercise, the tangential acceleration function:\[ a_{\text{tan}} = b + ct^2 \] Being given the condition that at time \(t=0\), the velocity \(v = v_0\), we integrate the acceleration with respect to time to find velocity:\[ \int dv = \int (b + ct^2) \ dt \] Upon integration, we get:\[ v = v_0 + bt + \frac{ct^3}{3} \]This integration reveals a velocity function that encapsulates the initial velocity and the effects of increasing tangential acceleration over time with terms involving time to the first and third power. Integration transforms our understanding from a static view of instantaneous change to a more dynamic view of movement over time. This step is crucial in physics problems as it allows us to transition from isolated data points to a continuous function, offering a complete understanding of an object's motion in a circular path.