Question

A discus thrower (with arm length of \(1.2 \mathrm{~m}\) ) starts from rest and begins to rotate counterclockwise with an angular acceleration of \(2.5 \mathrm{rad} / \mathrm{s}^{2}\) a) How long does it take the discus thrower's speed to get to \(4.7 \mathrm{rad} / \mathrm{s} ?\) b) How many revolutions does the thrower make to reach the speed of \(4.7 \mathrm{rad} / \mathrm{s} ?\) c) What is the linear speed of the discus at \(4.7 \mathrm{rad} / \mathrm{s} ?\) d) What is the linear acceleration of the discus thrower at this point? e) What is the magnitude of the centripetal acceleration of the discus thrown? f) What is the magnitude of the discus's total acceleration?

Short answer

Question: Calculate the magnitude of the total acceleration of the discus when it reaches an angular velocity of \(4.7 rad/s\), given that the angular acceleration is \(2.5 rad/s^2\) and the length of the thrower's arm is \(1.2m\). Answer: The magnitude of the total acceleration of the discus is approximately \(26.55 m/s^2\).

Step-by-step solution

a) Time to reach the desired angular velocity

The angular acceleration α is given \(2.5 rad/s^2\). We have to find the time it takes to reach the angular velocity \(4.7 rad/s\). We can use the following equation of motion: ω = ω0 + αt Here, ω is the final angular velocity, ω0 is the initial angular velocity which is \(0\) (since the discus thrower starts from rest), α is the angular acceleration, and t is the time. So, \(4.7 = 0 + 2.5t\) Solving for t, we get \(t = \frac{4.7}{2.5} = 1.88s\)

b) Number of revolutions to reach the angular velocity

To find the number of revolutions, we first need to find out the angular displacement. We can use the following equation of motion: θ = ω0t + 0.5αt^2 θ = 0.5 * 2.5 * (1.88)^2 = 4.425 rad Now, we can find the number of revolutions (n) by dividing the angular displacement θ by \(2π\): n = θ / (2π) = 4.425 / (2π) ≈ 0.704 revolutions

c) Linear speed of the discus

The linear speed (v) is the product of the angular velocity (ω) and the arm length (r): v = ω * r = 4.7 * 1.2 = 5.64 m/s

d) Linear acceleration of the discus thrower

The linear acceleration (a) is the product of the angular acceleration (α) and the arm length (r): a = α * r = 2.5 * 1.2 = 3 m/s^2

e) Centripetal acceleration of the discus thrown

The centripetal acceleration (a_c) is the product of the square of the angular velocity (ω) and the arm length (r): a_c = ω^2 * r = (4.7)^2 * 1.2 = 26.51 m/s^2

f) Magnitude of the discus's total acceleration

The total acceleration (a_total) is the sum of both the linear and centripetal accelerations squared and then taking the square root of that sum: a_total = sqrt(a^2 + a_c^2) = sqrt(3^2 + 26.51^2) ≈ 26.55 m/s^2

Key concepts

Angular Acceleration

Angular acceleration plays a crucial role in rotational motion, and its profound effect on an object's angular velocity can be seen in the example of a discus thrower. When the discus thrower spins, starting from rest, they experience angular acceleration, which is a measure of how quickly the angular velocity changes with time. In our scenario, the thrower has an angular acceleration of \(2.5 \text{rad/s}^2\).

This figure (\

Angular Displacement

Angular displacement is another key concept and describes the angle through which a point or line has been rotated in a specified sense about a specified axis. When calculating how many revolutions our discus thrower makes to reach a particular speed, we are essentially finding the angular displacement. By applying the motion equation \(\theta = 0.5 \cdot \alpha \cdot t^2\), we can find that the thrower moves through approximately 4.425 radians. This helps us understand the thrower's rotational path by converting it into a more relatable measure, revolutions, where one revolution equals \(2\pi\) radians.

By understanding angular displacement, students can better visualize the rotational equivalent of distance traveled in linear motion.

Linear Speed

Exploring the relationship between rotational and linear motion, linear speed illustrates how fast a point is moving along a path. It can be calculated by multiplying the angular velocity by the radius of the rotational motion. For the discus thrower with an arm length of 1.2 meters, and an angular velocity of \(4.7 \text{rad/s}\), their linear speed at the discus would be \(5.64 \text{m/s}\).

This concept bridges the gap between circular and straight-line motion, enabling students to see the direct connection between an object's speed along a path (linear speed) and its speed around a circle (angular velocity).

Centripetal Acceleration

Lastly, centripetal acceleration is the inward acceleration required to keep an object moving in a circle at a constant speed. It's a crucial component to understanding circular motion. In the case of our exercise, the discus experiences a centripetal acceleration because it is being whirled at the end of the thrower's arm. This acceleration can be calculated using the square of the angular velocity multiplied by the radial distance from the center of rotation. With the given values, the discus' centripetal acceleration reaches a significant magnitude of \(26.51 \text{m/s}^2\).

Understanding centripetal acceleration helps students grasp why an object in circular motion doesn't 'fly away' from the circle, and instead what forces are in play to keep it moving in a curved path.