
A disc has a diameter of one meter and rotates about an axis passing through its center and at right angles to its plane at the rate of 120 rev/min. Calculate the angular and linear velocities of a point on the rim and at a point halfway to the center?
Answer
550.8k+ views
Hint: The linear velocity is defined as the rate of change of the displacement to the time. Angular velocity is defined as the rate of change of velocity of an object when it is rotating around a point.
Complete step by step solution:
Given data:
The diameter of the disc = 1 m
Then the radius of the disc = 0.5 m
N = 120 rev/min
Let ${r_p}$ be a point halfway to the center and thus ${r_p} = \dfrac{{0.5}}{2} = 0.25m$
Let $\omega $ be the angular speed of the point on the rim. We know that $\omega $in terms of N is given by the formula,
$\omega = \dfrac{{2\pi N}}{{60}}$
$ \Rightarrow \omega = \dfrac{{2\pi N}}{{60}} \Rightarrow \dfrac{{2 \times \pi \times 120}}{{60}}$
$ \Rightarrow \omega = 4\pi = 12.56rad/s$
Let ${v_T}$be the linear speed of a point on the rim. We know that ${v_T}$in terms of $\omega ,$ given by the formula,
$\omega = \dfrac{{{v_T}}}{{{t_T}}}$
\[ \Rightarrow {v_T} = \omega {t_T}\]
\[ \Rightarrow {v_T} = 12.56 \times 0.5 = 6.28m/s\]
Similarly, let ${v_p}$be the linear velocity of a point halfway. Then ${v_p}$is given by the formula,
${v_p} = {r_p}\omega = 0.25 \times 12.56 = 3.14m/s$$\left( {\because v = r\omega } \right)$
$\Rightarrow $The linear velocity of the point on the rim $ = 6.28m/s$
$\Rightarrow $The angular velocity of the point on the rim $ = 12.56rad/s$
$\Rightarrow $The linear velocity of the point halfway $ = 3.14m/s$
$\Rightarrow $The angular velocity of the point halfway $ = 12.56rad/s$.
Note: 1. In a uniform circular motion, the angular velocity is equal to the angular displacement divided by the change in time.
2. Angular speed is a scalar quantity and will have the only magnitude whereas the angular velocity will have both the direction and the magnitude.
3. The angular velocity will be greater when the rotation angle is greater in the given interval of time. In a circular path, linear velocity increases when the radius increases and decreases when the radius decreases.
Complete step by step solution:
Given data:
The diameter of the disc = 1 m
Then the radius of the disc = 0.5 m
N = 120 rev/min
Let ${r_p}$ be a point halfway to the center and thus ${r_p} = \dfrac{{0.5}}{2} = 0.25m$
Let $\omega $ be the angular speed of the point on the rim. We know that $\omega $in terms of N is given by the formula,
$\omega = \dfrac{{2\pi N}}{{60}}$
$ \Rightarrow \omega = \dfrac{{2\pi N}}{{60}} \Rightarrow \dfrac{{2 \times \pi \times 120}}{{60}}$
$ \Rightarrow \omega = 4\pi = 12.56rad/s$
Let ${v_T}$be the linear speed of a point on the rim. We know that ${v_T}$in terms of $\omega ,$ given by the formula,
$\omega = \dfrac{{{v_T}}}{{{t_T}}}$
\[ \Rightarrow {v_T} = \omega {t_T}\]
\[ \Rightarrow {v_T} = 12.56 \times 0.5 = 6.28m/s\]
Similarly, let ${v_p}$be the linear velocity of a point halfway. Then ${v_p}$is given by the formula,
${v_p} = {r_p}\omega = 0.25 \times 12.56 = 3.14m/s$$\left( {\because v = r\omega } \right)$
$\Rightarrow $The linear velocity of the point on the rim $ = 6.28m/s$
$\Rightarrow $The angular velocity of the point on the rim $ = 12.56rad/s$
$\Rightarrow $The linear velocity of the point halfway $ = 3.14m/s$
$\Rightarrow $The angular velocity of the point halfway $ = 12.56rad/s$.
Note: 1. In a uniform circular motion, the angular velocity is equal to the angular displacement divided by the change in time.
2. Angular speed is a scalar quantity and will have the only magnitude whereas the angular velocity will have both the direction and the magnitude.
3. The angular velocity will be greater when the rotation angle is greater in the given interval of time. In a circular path, linear velocity increases when the radius increases and decreases when the radius decreases.
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