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Chapter 5: Problem 17

A point on the periphery of a rotating disc has its acceleration vector making an angle of \(30^{\circ}\) with the velocity vector. The ratio \(a_{c} / a_{t}\left(a_{c}\right.\) is centripetal acceleration and \(a_{t}\) is tangential acceleration) equals: (a) \(\sin 30^{\circ}\). (b) \(\cos 30^{\circ}\) (c) \(\tan 30^{\circ}\) (d) none of these

Short Answer

Expert verified
The ratio \(\frac{a_{c}}{a_{t}}\) equals \(\tan 30^{\circ}\), option (c).

Step by step solution

01

Understanding the problem

We have a point on a rotating disc. The acceleration vector forms an angle of \(30^{\circ}\) with the velocity vector. We need to find the ratio \(\frac{a_{c}}{a_{t}}\), where \(a_{c}\) is the centripetal acceleration and \(a_{t}\) is the tangential acceleration.
02

Express the acceleration in vector components

Since the total acceleration makes an angle of \(30^{\circ}\) with the velocity, we can use trigonometry to express \(a_{c}\) and \(a_{t}\). The total acceleration \(a\) can be broken down as: \(a \cos(30^{\circ}) = a_{c}\) and \(a \sin(30^{\circ}) = a_{t}\).
03

Setting up the ratio

The given problem asks for the ratio \(\frac{a_{c}}{a_{t}}\). Using the trigonometric expressions from the previous step, we have \(\frac{a \cos(30^{\circ})}{a \sin(30^{\circ})}\). Simplifying, this is \(\frac{\cos(30^{\circ})}{\sin(30^{\circ})}\).
04

Calculate the ratio

We have \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\) and \(\sin(30^{\circ}) = \frac{1}{2}\). Therefore, \(\frac{\cos(30^{\circ})}{\sin(30^{\circ})} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\).
05

Select the correct choice

The ratio \(\frac{a_{c}}{a_{t}} = \sqrt{3}\), which corresponds to \(\tan 30^{\circ}\). Thus, option (c) is correct.

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is a key concept when dealing with rotational motion. It acts towards the center of the circular path. Every object moving in a circle experiences this inward-directed force to keep it along its path. Think of it like a string keeping an object tethered as it spins around. If there wasn't centripetal acceleration, the object would move off in a straight line due to inertia.

Mathematically, centripetal acceleration (\(a_{c}\)) can be expressed as:\[a_{c} = \frac{v^2}{r}\]Here, \(v\) represents the tangential speed, and \(r\) is the radius of the circular path.

This relationship shows that as the speed of an object increases or if it spins along a smaller radius, the centripetal acceleration will increase. It's crucial to remember that centripetal force is required for this acceleration and is often provided by gravitational, tension, or frictional forces depending on the situation.
Tangential Acceleration
Tangential acceleration is somewhat different from centripetal acceleration. While centripetal acceleration changes an object's direction, tangential acceleration changes its speed along the circular path. Imagine accelerating in a car around a roundabout: while you are turning, if you press the gas pedal, you feel a surge forward due to tangential acceleration.

Tangential acceleration (\(a_{t}\)) is related to how fast the speed of an object changes as it moves along the curve:\[a_{t} = \frac{dv}{dt}\]where \(dv\) is the change in velocity and \(dt\) is the change in time.

In problems involving rotational motion, both centripetal and tangential accelerations can be present and contribute to an object's total acceleration. The total acceleration is then a combination of these two and can be decomposed into components using trigonometric methods.
Trigonometric Ratios
Trigonometric ratios are incredibly useful in understanding rotational motion, especially when determining the components of vector quantities like acceleration. In the context of our exercise, we use these ratios to decompose the total acceleration into centripetal and tangential components.

Consider a scenario where an acceleration vector makes an angle \(\theta\) with a velocity vector. Here鈥檚 how we break it down:
  • The component parallel to the direction of velocity (tangential component) would be \(a_t = a \sin(\theta)\)
  • The component perpendicular to the velocity (centripetal component) would be \(a_c = a \cos(\theta)\)
In our problem, the acceleration vector forms a \(30^{\circ}\) angle with the velocity vector. We use the trigonometric identities for \(\sin\) and \(\cos\) of \(30^{\circ}\) to find the relationship between the accelerations.
Ultimately, the ratio derived is \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\), which is used to determine how these components relate quantitatively.

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Most popular questions from this chapter

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Three identical cars \(A, B\) and \(C\) are moving at the same speed on three bridges. The car \(A\) goes on a plane bridge, \(B\) on a bridge convex upwards and \(C\) goes on a bridge concave upwards. Let \(F_{A}, F_{B}\) and \(F_{C}\) be the normal forces exerted by the cars on the bridges when they are at the middle of the bridges. Then: (a) \(F_{A}\) is maximum of the three forces (b) \(F_{B}\) is maximum of the three forces (c) \(F_{C}\) is maximum of the three forces (d) \(F_{A}=F_{B}=F_{C}\)
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