A particle of unit mass is moving on a plane. Its trajectory, in polar coordinates, is given by r(t) = t2, θ(t) = t where t is time. The kinetic energy of the particle at time t = 2 is
16
Understanding the motion of a particle in polar coordinates requires calculating its velocity components to determine its kinetic energy. This problem involves a particle with unit mass and a given trajectory in polar coordinates, \(r(t)\) and \(\theta(t)\).
Particle Motion Analysis
The problem provides the following information about the particle:
- Mass of the particle (m): Unit mass, so \(m = 1\).
- Radial position (r): Given as a function of time, \(r(t) = t^2\).
- Angular position (\(\theta\)): Given as a function of time, \(\theta(t) = t\).
- Time (t): We need to find the kinetic energy at \(t = 2\).
To calculate the kinetic energy, we first need to determine the particle's velocity. In polar coordinates, the square of the velocity (\(v^2\)) is given by the formula:
$$v^2 = \dot{r}^2 + (r\dot{\theta})^2$$
where \(\dot{r}\) is the radial velocity (rate of change of radial position) and \(\dot{\theta}\) is the angular velocity (rate of change of angular position).
Velocity Components in Polar Coordinates
Let's calculate the time derivatives of the given radial and angular positions to find the velocity components:
- Radial Velocity (\(\dot{r}\)):
Given \(r(t) = t^2\), we differentiate with respect to time (t):
$$\dot{r}(t) = \frac{dr}{dt} = \frac{d}{dt}(t^2) = 2t$$
- Angular Velocity (\(\dot{\theta}\)):
Given \(\theta(t) = t\), we differentiate with respect to time (t):
$$\dot{\theta}(t) = \frac{d\theta}{dt} = \frac{d}{dt}(t) = 1$$
Values at Time \(t = 2\)
Now, we substitute the specific time \(t = 2\) into the expressions for \(r(t)\), \(\dot{r}(t)\), and \(\dot{\theta}(t)\) to find their values at that instant:
- \(r(2) = (2)^2 = 4\)
- \(\dot{r}(2) = 2(2) = 4\)
- \(\dot{\theta}(2) = 1\) (since it's a constant)
Kinetic Energy Calculation
With the values of \(r\), \(\dot{r}\), and \(\dot{\theta}\) at \(t = 2\), we can calculate the square of the particle's velocity (\(v^2\)) using the polar coordinates velocity formula:
$$v^2 = (\dot{r})^2 + (r\dot{\theta})^2$$
Substitute the values at \(t = 2\):
$$v^2 = (4)^2 + (4 \times 1)^2$$
$$v^2 = 16 + (4)^2$$
$$v^2 = 16 + 16$$
$$v^2 = 32$$
Finally, we use the standard formula for kinetic energy:
$$KE = \frac{1}{2}mv^2$$
Substitute the mass \(m = 1\) and the calculated \(v^2 = 32\):
$$KE = \frac{1}{2}(1)(32)$$
$$KE = 16$$
Thus, the kinetic energy of the particle at time \(t = 2\) is 16.
| Parameter | Expression/Given | Value at \(t = 2\) |
|---|---|---|
| Mass (m) | Unit mass | 1 |
| Radial position (r) | \(r(t) = t^2\) | \(r(2) = 2^2 = 4\) |
| Radial velocity (\(\dot{r}\)) | \(\dot{r}(t) = 2t\) | \(\dot{r}(2) = 2(2) = 4\) |
| Angular position (\(\theta\)) | \(\theta(t) = t\) | \(\theta(2) = 2\) |
| Angular velocity (\(\dot{\theta}\)) | \(\dot{\theta}(t) = 1\) | \(\dot{\theta}(2) = 1\) |
| Velocity squared (\(v^2\)) | \(v^2 = \dot{r}^2 + (r\dot{\theta})^2\) | \(4^2 + (4 \times 1)^2 = 16 + 16 = 32\) |
| Kinetic Energy (KE) | \(KE = \frac{1}{2}mv^2\) | \(\frac{1}{2}(1)(32) = 16\) |
Important Questions from Work Power and Energy
Which of the following is an example of potential energy?
Energy in natural systems is transferred by _______.
Choose the best option from the following.
_______ is one of the chemical used to produce Ocean Thermal Energy.
A body of mass 2kg is thrown up vertically with a kinetic energy of 500J. If the acceleration due to gravity is 10 m/s2, the height at which the kinetic energy of the body becomes half of the original values is _____
Arjun has a mass of 50 kg, he eats banana of 1000 calories. If this energy is used up to lift him from the ground, the height to which he can climb is _______. Take g = 10 m/s2 and 1 calories = 4.2 Joule.
