/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 A point \(P\) moves in counter-c... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 85

A point \(P\) moves in counter-clockwise direction on a circular path as shown in Fig. 6.44. The movement of \(P\) is such that it sweeps out a length \(s=t^{3}+5\), where s is in metres and \(t\) is in seconds. The radius of the path is \(20 \mathrm{~m}\). The acceleration of \(P\) when \(t=2\) is nearly (A) \(13 \mathrm{~m} / \mathrm{s}^{2}\) (B) \(12 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(7.2 \mathrm{~m} / \mathrm{s}^{2}\) (D) \(14 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The acceleration of point $P$ when $t=2$ seconds is nearly \(14 \mathrm{~m} / \mathrm{s}^{2}\) (Option D).

Step by step solution

01

Find the expression for velocity

To find the velocity expression, differentiate the given displacement equation (s=t^3+5) with respect to time t: v = ds/dt = d(t^3 + 5)/dt = 3t^2
02

Find the expression for acceleration

To find the acceleration expression, differentiate the obtained velocity expression (v=3t^2) with respect to time t: a = dv/dt = d(3t^2)/dt = 6t
03

Find the centripetal acceleration

Centripetal acceleration is the acceleration of an object moving in a circle at a constant speed. The formula for centripetal acceleration is: ac = v^2/R where R is the radius of the circular path and v is the tangential velocity. Given R=20 m, we need to find the centripetal acceleration at t=2 seconds.
04

Calculate velocities and centripetal acceleration at t=2 seconds

Substitute t=2 seconds into the expression for tangential velocity v=3t^2: v(2) = 3(2^2) = 12 m/s Now we can find the centripetal acceleration at t=2 seconds using the formula ac=v^2/R: ac(2) = (12 m/s)^2 / (20 m) = 7.2 m/s^2
05

Calculate total acceleration at t=2 seconds

Now we have centripetal acceleration and the expression for tangential acceleration. At t=2 seconds, tangential acceleration can be determined using the expression a=6t: a(2) = 6(2) = 12 m/s^2 The total acceleration is the vector sum of both tangential and centripetal accelerations. We can find the magnitude of total acceleration using the Pythagorean theorem: a_total = sqrt(a^2 + ac^2) Substitute the values of tangential and centripetal accelerations at t=2 seconds: a_total(2) = sqrt((12 m/s^2)^2 + (7.2 m/s^2)^2) 鈮� 13.93 m/s^2
06

Compare the calculated total acceleration with given options

Comparing the calculated total acceleration (a_total(2) 鈮� 13.93 m/s^2) with the given options: (A) 13 m/s^2 - Not close enough (B) 12 m/s^2 - Not close enough (C) 7.2 m/s^2 - Not close enough (D) 14 m/s^2 - Close enough Hence, the correct answer is option (D) 14 m/s^2.

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Key Concepts

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

Tangential Acceleration
Tangential acceleration refers to the rate of change of the tangential velocity of an object moving along a circular path.
It is directly related to the speed at which the object moves around the circle's edge.
In the given problem, we see that tangential acceleration changes based on the time variable, due to the relationship given by the displacement equation.
  • The expression for tangential velocity was found by differentiating the displacement: \( v = \frac{ds}{dt} = 3t^2 \).
  • Next, the tangential acceleration was obtained by differentiating the velocity: \( a = \frac{dv}{dt} = 6t \).
When \( t = 2 \), the tangential acceleration becomes \( a(2) = 6 \times 2 = 12 \text{ m/s}^2 \).
This acceleration acts in the direction of the motion on the circle, increasing or decreasing the speed as the object moves along its path.
Centripetal Acceleration
Centripetal acceleration is a key concept involved when an object travels in a circular motion.
It always points towards the center of the circle. This type of acceleration is essential to keep the object moving around the circle rather than flying off in a straight line.
  • The formula for centripetal acceleration is: \( a_c = \frac{v^2}{R} \), where \( v \) is the tangential velocity and \( R \) is the radius of the circular path.
  • Using the velocity calculated earlier, when \( t = 2 \), we find \( v(2) = 12 \text{ m/s} \).
  • Given the radius \( R = 20 \text{ m} \), the centripetal acceleration at \( t=2 \) becomes \( a_c(2) = \frac{(12 \text{ m/s})^2}{20 \text{ m}} = 7.2 \text{ m/s}^2 \).
This centripetal acceleration is crucial as it makes sure the object remains on its circular path, tethered towards the center.
Differentiation in Physics
Differentiation is a mathematical tool widely used in physics to describe how physical quantities change with respect to time or other variables.
In physics problems involving motion, such as this problem, differentiation helps calculate velocity and acceleration from a displacement-time equation.
  • In the exercise, we started with a displacement function: \( s = t^3 + 5 \).
  • The first derivative with respect to time (\( t \)) gave the velocity: \( v = ds/dt = 3t^2 \).
  • A second derivative provided the expression for the tangential acceleration: \( a = dv/dt = 6t \).
By differentiation, one can discern how quickly an object accelerates or decelerates, or how fast it is moving at any given time.
Understanding differentiation is like unlocking the rules that govern motion dynamics, allowing us to accurately describe and predict motion.

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Most popular questions from this chapter

A round uniform body of radius \(R\), mass \(m\) and moment of inertia \(I\) rolls down (without slipping) an inclined plane making an angle with the horizontal. Then its acceleration is \(\quad\) [2007] (A) \(\frac{g \sin \theta}{1+\frac{m R^{2}}{I}}\) (B) \(\frac{g \sin \theta}{1-\frac{I}{m R^{2}}}\) (C) \(\frac{g \sin \theta}{1-\frac{m R^{2}}{I}}\) (D) \(\frac{g \sin \theta}{1+\frac{I}{m R}}\)

A hoop of radius \(R\) and mass \(m\) rotating with an angular velocity \(\omega_{0}\) is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity if the centre of the loop ceases to slip? (A) \(\frac{r \omega_{0}}{3}\) (B) \(\frac{r \omega_{0}}{2}\) (C) \(r \omega_{0}\) (D) \(\frac{r \omega_{0}}{4}\)

A horizontal disc rotates freely about a vertical axis through its centre. A ring, having the same mass and radius as the disc, is now gently placed on the disc. After some time, the two rotate with a common angular velocity (A) some friction exists between the disc and the ring. (B) the angular momentum of the disc plus ring is conserved. (C) the final common angular velocity is \(\frac{2}{3} \mathrm{rd}\) of the initial angular velocity of the disc. (D) \(\frac{2}{3}\) rd of the initial kinetic energy changes to heat,

Four point masses, each of value \(m\), are placed at the corners of a square \(\mathrm{ABCD}\) of side \(\ell\). The moment of inertia of this system about an axis passing through \(A\) and parallel to BD iis (A) \(3 m \ell^{2}\) (B) \(m \ell^{2}\) (C) \(2 m \ell^{2}\) (D) \(\sqrt{3} m \ell^{2}\)

The moment of inertia of a circular wire of mass \(m\) and radius \(R\) about its diameter is (A) \(m R^{2} / 2\) (B) \(m R^{2}\) (C) \(2 m R^{2}\) (D) \(m R^{2} / 4\)
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