In the act of jumping off a diving board, a diver changed his angular velocity from zero to \(6.20 \mathrm{rad} / \mathrm{s}\) in \(220 \mathrm{~ms}\). The diver's rotational inertia is \(12.0 \mathrm{~kg} \cdot \mathrm{m}^{2} .(a)\) Find the angular acceleration during the jump. (b) What external torque acted on the diver during the jump?

Short Answer

Expert verified
The angular acceleration during the jump is 28.18 rad/s^2. The external torque acted on the diver during the jump is 338.16 N.m.

Step by step solution

01

Calculate Angular Acceleration

Angular acceleration (denoted by the Greek Letter alpha \(\alpha\)) is calculated using the formula: \(\alpha = (w_f - w_i) / t\). In our problem, \(w_f = 6.20 \ rad/s\), which is the final angular velocity, \(w_i = 0 \ rad/s\), the initial angular velocity and \(t = 220 \ ms = 0.22 \ s\). Substituting these values into the formula, and solving for \(\alpha\), we have \(\alpha = (6.20 - 0) / 0.22 = 28.18 \ rad/s^2\).
02

Calculate Torque

Torque (Symbol T) is calculated using the formula: \(T = I \cdot \alpha\), where \(I = 12.0 \ kg \cdot m ^ {2}\) is the moment of inertia, and \(\alpha = 28.18 \ rad / s^2\) is the angular acceleration we calculated in the previous step. Substituting these values into the formula, and solving for T, We have \(T = 12.0 \cdot 28.18 = 338.16 \ N \cdot m\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque
Torque is a measure of the force that can cause an object to rotate about an axis. Simply put, it's like a twist or a turn that can make something spin. It's a vital concept in physics because it helps us understand how objects rotate and how much force is needed to change their rotational state.

Think of torque as the power you need to use to open a tight jar lid - your hands apply a force at a distance from the center of the lid, causing it to rotate. In physics, we calculate torque (\(T\)) using the equation: \[ T = I \times \alpha \] where \(I\) is the moment of inertia (which measures an object's resistance to changes in its rotation), and \(\alpha\) is the angular acceleration.

Using the exercise as an example, to find the total torque exerted by the diver, we use his moment of inertia and the angular acceleration we calculated. This torque is the product of the two, indicating how strongly the diver needed to 'push off' to achieve the rotation.
Moment of Inertia
The moment of inertia (\(I\) is a bit like the mass in rotational motion. It's a property of an object that tells us how difficult it is to change its rotational speed. Different objects will have different moments of inertia, even if they have the same mass, depending on how that mass is distributed relative to the axis of rotation.

For instance, a figure skater pulling in their arms reduces their moment of inertia, making it easier to spin faster. This conservation of angular momentum is fundamental in many areas of physics and engineering.

In the diver's case, a moment of inertia of \(12.0 \mathrm{kg} \cdot \mathrm{m}^{2}\) indicates how his body's mass distribution affects his ability to rotate as he jumps off the diving board. The lower the moment of inertia, the easier it is to achieve higher angular acceleration for a given torque.
Angular Velocity
Angular velocity is how fast something is rotating, or you can think of it as the rotational equivalent of linear velocity. When you're looking at something that spins - like a wheel, a planet, or a diver mid-jump - angular velocity tells you the rate at which they're turning.

In our example, the diver's angular velocity increased from \(0 \mathrm{rad/s}\) (not rotating) to \(6.20 \mathrm{rad/s}\) (rotating quite fast!). Calculating this change gives us the angular acceleration, which we can then use to determine the required torque, as seen in the exercise. The relationship between torque and angular velocity is also why engines are rated with horsepower and torque - because it tells you how quickly they can make the wheels turn and how potent that turn can be.

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Most popular questions from this chapter

What minimum force \(F\) applied horizontally at the axle of the wheel in Fig. \(9-49\) is necessary to raise the wheel over an obstacle of height \(h\) ? Take \(r\) as the radius of the wheel and \(W\) as its weight.

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