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Chapter 5: Problem 63

(III) A particle rotates in a circle of radius \(3.80 \mathrm{~m}\). At a particular instant its acceleration is \(1.15 \mathrm{~m} / \mathrm{s}^{2}\) in a direction that makes an angle of \(38.0^{\circ}\) to its direction of motion. Determine its speed \((a)\) at this moment and \((b) 2.00 \mathrm{~s}\) later, assuming constant tangential acceleration.

Short Answer

Expert verified
Speed initially is approximately 1.86 m/s; after 2 seconds, it's 3.28 m/s.

Step by step solution

01

Understand the Given Information

We have a particle moving in a circle of radius \( r = 3.80 \text{ m} \). The particle's acceleration is \( 1.15 \text{ m/s}^2 \), which is at an angle of \( 38.0^{\circ} \) to its direction of motion. We need to find (a) the speed at this instant and (b) the speed 2 seconds later with constant tangential acceleration.
02

Resolve Acceleration into Components

The given acceleration can be resolved into two components: radial acceleration (\( a_r \)) and tangential acceleration (\( a_t \)). The radial acceleration is given by \( a \cos(38.0^{\circ}) \) and the tangential acceleration by \( a \sin(38.0^{\circ}) \), where \( a = 1.15 \text{ m/s}^2 \).
03

Calculate Radial Acceleration

Calculate the radial component of the acceleration: \[ a_r = a \cos(38.0^{\circ}) = 1.15 \times \cos(38.0^{\circ}) \approx 0.91 \text{ m/s}^2 \] This radial component is also equal to \( \frac{v^2}{r} \), where \( v \) is the speed we want to find.
04

Solve for Initial Speed

Set the radial acceleration equal to \( \frac{v^2}{r} \) to solve for the speed:\[ 0.91 = \frac{v^2}{3.80} \]Multiply both sides by 3.80 and solve for \( v \): \[ v^2 = 3.80 \times 0.91 \]\[ v^2 = 3.458 \]\[ v = \sqrt{3.458} \approx 1.86 \text{ m/s} \].
05

Calculate Tangential Acceleration

Calculate the tangential component of the acceleration: \[ a_t = a \sin(38.0^{\circ}) = 1.15 \times \sin(38.0^{\circ}) \approx 0.71 \text{ m/s}^2 \].This tangential acceleration is constant and will be used to calculate the speed after 2 seconds.
06

Calculate Speed After 2 Seconds

Using the formula for speed with constant tangential acceleration: \[ v_f = v_i + a_t \times t \]where \( v_i = 1.86 \text{ m/s} \) and \( t = 2 \text{ s} \), \[ v_f = 1.86 + 0.71 \times 2 \]\[ v_f = 1.86 + 1.42 = 3.28 \text{ m/s} \].

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Key Concepts

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

Tangential Acceleration
When an object is rotating in a circle, its velocity isn't constant because the direction is constantly changing. However, if there is also a change in speed along the circular path, this is due to tangential acceleration.
The tangential acceleration, often denoted as \(a_t\), affects how fast or slow an object speeds up or slows down while moving along the circular path. This is analogous to the acceleration in linear motion, just along the tangent to the circular path.
  • The formula for tangential acceleration when the component of total acceleration and the angle is known is: \\[a_t = a \sin(\theta)\]
  • In this exercise, where \(a = 1.15 \, \text{m/s}^2\) and \(\theta = 38.0^\circ\), we find \(a_t\) to be approximately \(0.71 \, \text{m/s}^2\).
  • It is constant in this problem, allowing the use of simple kinematic equations to find the change in speed over time.
Radial Acceleration
Another critical component when dealing with circular motion is radial acceleration, which is directed towards the center of the circular path.
This inward acceleration keeps the particle moving in a circle rather than flying off in a straight line. Radial acceleration is also known as centripetal acceleration and is given by the formula \[a_r = \frac{v^2}{r}\]where \(v\) is the speed of the particle and \(r\) is the radius of the circle.
  • It stems from the component of acceleration that points towards the center, calculated as: \\[a_r = a \cos(\theta)\]
  • For the particle in the problem: with \(a = 1.15 \, \text{m/s}^2\) and \(\theta = 38.0^\circ\), the radial acceleration \(a_r\) equals approximately \(0.91 \, \text{m/s}^2\).
  • It ensures the particle maintains its circular path by constantly changing the direction of the velocity.
Speed Calculation
Calculating speed in circular motion can involve several steps depending on if we are finding the speed at one instant or after a period, accounting for tangential acceleration.
The initial speed \(v\) can be solved by equating the radial acceleration to the centripetal formula: \[a_r = \frac{v^2}{r}\]From this, in the exercise, we solved for \(v\) to find it was \(1.86 \,\text{m/s}\).
  • Once the initial speed \(v_i\) is determined, future speed can be calculated using the formula for constant acceleration: \\[v_f = v_i + a_t \times t\]
  • Since the tangential acceleration \(a_t\) is constant, the final speed after 2 seconds, as calculated in the exercise, becomes \(3.28 \,\text{m/s}\).
  • This process highlights how both types of acceleration impact speed: radial acceleration affects direction, while tangential acceleration affects magnitude.

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Most popular questions from this chapter

(III) A curve of radius \(68 \mathrm{~m}\) is banked for a design speed of \(85 \mathrm{~km} / \mathrm{h}\). If the coefficient of static friction is 0.30 (wet pavement), at what range of speeds can a car safely make the curve? [Hint: Consider the direction of the friction force when the car goes too slow or too fast.]

(II) The design of a new road includes a straight stretch that is horizontal and flat but that suddenly dips down a steep hill at \(22^{\circ} .\) The transition should be rounded with what minimum radius so that cars traveling \(95 \mathrm{~km} / \mathrm{h}\) will not leave the road (Fig. \(5-45) ?\)

(III) An object of mass \(m\) is constrained to move in a circle of radius \(r\). Its tangential acceleration as a function of time is given by \(a_{\tan }=b+c t^{2},\) where \(b\) and \(c\) are constants. If \(v=v_{0}\) at \(t=0,\) determine the tangential and radial components of the force, \(F_{\text {tan }}\) and \(F_{\mathrm{R}}\), acting on the object at any time \(t>0\).

(III) A 3.0 -kg block sits on top of a \(5.0-\mathrm{kg}\) block which is on a horizontal surface. The 5.0 -kg block is pulled to the right with a force \(\vec{\mathbf{F}}\) as shown in Fig. \(39 .\) The coefficient of static friction between all surfaces is 0.60 and the kinetic coeffi- cient is \(0.40 .\) (a) What is the minimum value of \(F\) needed to move the two blocks? (b) If the force is 10\(\%\) greater than your answer for \((a),\) what is the acceleration of each block?

(III) The position of a particle moving in the \(x y\) plane is given by \(\overrightarrow{\mathbf{r}}=2.0 \cos (3.0 \mathrm{rad} / \mathrm{s} t) \hat{\mathbf{i}}+2.0 \sin (3.0 \mathrm{rad} / \mathrm{s} t) \hat{\mathbf{j}}\) where \(r\) is in meters and \(t\) is in seconds. ( \(a\) ) Show that this represents circular motion of radius \(2.0 \mathrm{~m}\) centered at the origin. (b) Determine the velocity and acceleration vectors as functions of time. ( \(c\) ) Determine the speed and magnitude of the acceleration. \((d)\) Show that \(a=v^{2} / r .(e)\) Show that the acceleration vector always points toward the center of the circle.
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