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Chapter 7: Problem 7

The angular position \(\theta\) of any point \(P\) situated at a distanee \(r=2 \mathrm{~cm}\) from the axis of rotation of a whecl of radius \(R=10 \mathrm{~cm}\) moves with non-uniform cireular motion obcying the relation \(\theta=2 t^{3}-3 t^{2}+6\) in radians. Find the (a) angular velocity when \(\alpha=0\). (b) ratio of magnitudes of tangential and radial acceleration of the point \(P\).

Short Answer

Expert verified
(a) -3/2 rad/s (b) 0

Step by step solution

01

Differentiate Angular Position for Velocity

The angular velocity \( \omega \) is the derivative of angular position \( \theta \) with respect to time \( t \). Start from the given formula:\[\theta = 2t^3 - 3t^2 + 6\]Calculate \( \omega = \frac{d\theta}{dt} \):\[\omega = \frac{d}{dt}(2t^3 - 3t^2 + 6) = 6t^2 - 6t\]
02

Set Angular Acceleration to Zero

Given that the angular acceleration \( \alpha = 0 \), we know that acceleration is the derivative of velocity.\[\alpha = \frac{d\omega}{dt} \]Calculating \( \alpha \) yields:\[\alpha = \frac{d}{dt}(6t^2 - 6t) = 12t - 6\]Set \( \alpha = 0 \) to find the time \( t \):\[12t - 6 = 0 \Rightarrow t = \frac{1}{2}\]
03

Compute Angular Velocity at t = 0.5s

Substitute \( t = \frac{1}{2} \) into the angular velocity equation from Step 1:\[\omega = 6\left(\frac{1}{2}\right)^2 - 6\left(\frac{1}{2}\right) = \frac{3}{2} - 3 = -\frac{3}{2} \, \text{rad/s}\]
04

Determine Tangential Acceleration

Tangential acceleration \( a_t \) is given by:\[a_t = r \alpha\]Here \( r = 2 \) cm, \( \alpha = 0 \):Since \( \alpha = 0 \), \( a_t = 0 \).
05

Calculate Radial Acceleration

Radial acceleration \( a_r \) is given by:\[a_r = r \omega^2\]Substituting \( r = 2 \) cm and \( \omega = -\frac{3}{2} \) rad/s:\[a_r = 2\left(-\frac{3}{2}\right)^2 = 2 \times \frac{9}{4} = \frac{9}{2} \text{ cm/s}^2\]
06

Compute Ratio of Accelerations

The ratio of tangential to radial acceleration is:\[\text{Ratio} = \frac{|a_t|}{|a_r|}\]Where \( a_t = 0 \) and \( a_r = \frac{9}{2} \):\[\text{Ratio} = \frac{0}{\frac{9}{2}} = 0\]

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Key Concepts

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

Understanding Angular Velocity
Angular velocity tells us how quickly an object rotates around an axis. It is denoted by the Greek letter \( \omega \) and is measured in radians per second. To find angular velocity, we differentiate the angular position \( \theta \) with respect to time \( t \).
  • For a given rotation described by \( \theta = 2t^3 - 3t^2 + 6 \), the angular velocity is calculated as \( \omega = \frac{d\theta}{dt} = 6t^2 - 6t \).
  • This step involves taking the derivative of the polynomial in \( \theta \) with respect to time. Essentially, this provides a rate of change of the rotational angle, indicating how fast the rotation occurs at any time \( t \).
To find specific values like the angular velocity when the acceleration is zero, we set the derivative of angular velocity (which is angular acceleration) to zero and solve for \( t \).
  • After solving the equation \( 12t - 6 = 0 \) for \( t \), the result is \( t = \frac{1}{2} \) seconds. Substituting this time back into \( \omega = 6t^2 - 6t \) gives us the angular velocity at \( t = 0.5 \) seconds as \( \omega = -\frac{3}{2} \) rad/s.
Exploring Tangential Acceleration
Tangential acceleration is the rate at which the tangential velocity of a point on a rotating body changes. It is denoted by \( a_t \), and it acts along the direction of the tangent to the path followed by the rotating object.
  • In rotational motion, tangential acceleration is directly proportional to the angular acceleration \( \alpha \). The relation is given by \( a_t = r \alpha \), where \( r \) is the radius or the distance of the point from the axis of rotation.
  • In the scenario where \( \alpha = 0 \), as derived from the equation \( 12t - 6 = 0 \), the tangential acceleration \( a_t \) becomes \( 0 \) since there is no change in the angular velocity. This means the point moves with a constant rotational speed at the given instant.
Understanding this helps us determine how the speed of a point on a rotating path changes concerning rotation and is crucial in systems where rotational dynamics are considered.
Analyzing Radial Acceleration
Radial acceleration, also known as centripetal acceleration, pertains to the change in direction of velocity as an object moves in a circular path. It always points towards the center of the circle.
  • The radial acceleration \( a_r \) can be calculated using the formula \( a_r = r \omega^2 \), where \( \omega \) is the angular velocity, and \( r \) is the radius of rotation. This calculation involves squaring the angular velocity, reflecting its dependency on the square of rotational speed.
  • For our case, substituting \( r = 2 \text{ cm} \) and \( \omega = -\frac{3}{2} \text{ rad/s} \), we find \( a_r = 2 \times \left(-\frac{3}{2}\right)^2 = \frac{9}{2} \text{ cm/s}^2 \). This indicates a sustained acceleration towards the center, necessary to keep the object moving in its circular path.
This component of acceleration ensures that the object stays on its circular path rather than moving off in a tangent, demonstrating the continuous inward force required for circular motion.

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Most popular questions from this chapter

A solid sphere, hollow sphere, solid cylinder, hollow cylinder and ring cach of mass \(M\) and radius \(R\) are simultancously relcased at rest from top of inclinc and allowed to roll down the incline Column-I Column-II (a) Time taken to reach bottom is maximum for (p) Solid sphere (b) \(\Lambda\) ngular acceleration maximum for (q) 1 lollow cylinder (c) Kinctic cnergy at botom is same for (r) Hollow sphere (d) Rotational kinctic energy is maximum for (s) Ring

\(\Lambda\) uniform disc of mass \(m\) and radius \(R\) is rotated about an axis passing through its centre and perpendicular to its plane with an angular velocity \(\omega_{0} .\) It is placed on a rough horizontal plane with the axis of the disc keeping vertical. Coefficient of friction between the disc and the surface is \(\mu\). Find (a) the time when disc stops rotating, (b) the angle rotated by the disc before stopping.

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A whecl of radius \(r\) consists of a thin unifonn rim of muss \(m\) and six thin spokes of mass \(m\). What is the moment of incrtia of the whecl about axis passing through \(A\) and perpcndicular to the planc? (a) \(6 m r^{2}\) (b) \(4 m r^{2}\) (c) \(3 m^{2}\) (d) \(10 m r^{2}\)

A uniform thin rod of mass \(m\) and lenght \(l\) is standing on a smooth horizontal surface. \(\Lambda\) slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling. Find the velocity of centre of mass of the rod at the instant when it makes an angle \(\theta\) with horizontal.
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