Question

A particle is travelling in a circular path of radius 4m. At a certain instant the particle is moving at 20m/s and its acceleration is at an angle of 37º from the direction to the centre of the circle as seen from the particle

(i) At what rate is the speed of the particle increasing?

(ii) What is the magnitude of the acceleration?

Answer 1

(i) 75 m/s^2, (ii) 125 m/s^2

Answer 2

The problem describes a particle moving in a circular path. We are given its radius, instantaneous speed, and the angle of its total acceleration with respect to the direction pointing towards the center of the circle. We need to find the rate at which the speed is increasing (tangential acceleration) and the magnitude of the total acceleration.

The total acceleration ($\vec{a}$) of a particle in circular motion has two perpendicular components:

1. Centripetal acceleration ($\vec{a_c}$): This component is directed towards the center of the circle and is responsible for changing the direction of the velocity. Its magnitude is given by $a_c = \frac{v^2}{R}$.
2. Tangential acceleration ($\vec{a_t}$): This component is tangential to the circular path and is responsible for changing the magnitude of the velocity (speed). Its magnitude is given by $a_t = \frac{dv}{dt}$.

The magnitude of the total acceleration is given by $a = \sqrt{a_c^2 + a_t^2}$.

Given:

• Radius, $R = 4$ m
• Speed, $v = 20$ m/s
• Angle between total acceleration and the direction to the center, $\theta = 37^\circ$. This means the angle between $\vec{a}$ and $\vec{a_c}$ is $37^\circ$.

Step 1: Calculate the centripetal acceleration ($a_c$) $a_c = \frac{v^2}{R} = \frac{(20 \text{ m/s})^2}{4 \text{ m}} = \frac{400 \text{ m}^2/\text{s}^2}{4 \text{ m}} = 100 \text{ m/s}^2$

Step 2: Find the tangential acceleration ($a_t$) From the vector diagram of acceleration components, $a_c$ and $a_t$ are perpendicular. The angle $\theta = 37^\circ$ is between the total acceleration $\vec{a}$ and the centripetal acceleration $\vec{a_c}$. Using trigonometry: $\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{a_t}{a_c}$ We use the common approximation for $37^\circ$ from a 3-4-5 right triangle: $\tan(37^\circ) \approx \frac{3}{4} = 0.75$. $a_t = a_c \tan(37^\circ) = 100 \text{ m/s}^2 \times 0.75 = 75 \text{ m/s}^2$ This is the rate at which the speed of the particle is increasing.

Step 3: Calculate the magnitude of the total acceleration ($a$) We can use the Pythagorean theorem or trigonometry. Using trigonometry: $\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{a_c}{a}$ We use the common approximation for $37^\circ$: $\cos(37^\circ) \approx \frac{4}{5} = 0.8$. $a = \frac{a_c}{\cos(37^\circ)} = \frac{100 \text{ m/s}^2}{0.8} = \frac{1000}{8} \text{ m/s}^2 = 125 \text{ m/s}^2$ Alternatively, using $a = \sqrt{a_c^2 + a_t^2}$: $a = \sqrt{(100 \text{ m/s}^2)^2 + (75 \text{ m/s}^2)^2} = \sqrt{10000 + 5625} \text{ m/s}^2 = \sqrt{15625} \text{ m/s}^2 = 125 \text{ m/s}^2$

The final answer is $\boxed{\text{(i) 75 m/s}^2, \text{(ii) 125 m/s}^2}$.

Explanation of the solution:

1. Calculate centripetal acceleration $a_c = v^2/R = (20^2)/4 = 100 \text{ m/s}^2$.
2. Identify tangential acceleration $a_t$ as the rate of increase of speed. The angle $37^\circ$ is between the total acceleration $a$ and $a_c$.
3. Use $\tan(37^\circ) = a_t/a_c$. Since $\tan(37^\circ) \approx 0.75$, $a_t = 100 \times 0.75 = 75 \text{ m/s}^2$.
4. Calculate the magnitude of total acceleration $a$ using $\cos(37^\circ) = a_c/a$. Since $\cos(37^\circ) \approx 0.8$, $a = 100/0.8 = 125 \text{ m/s}^2$.

Answer:

(i) The rate at which the speed of the particle is increasing is $75 \text{ m/s}^2$. (ii) The magnitude of the acceleration is $125 \text{ m/s}^2$.