6.6 Non-uniform circular motion  (Page 4/4)

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$\begin{array}{l}⇒{a}_{T}=0\\ ⇒\alpha =0\end{array}$

Description of circular motion using vectors

In the light of new quantities and new relationships, we can attempt analysis of the general circular motion (including both uniform and non-uniform), using vector relations. We have seen that :

$\begin{array}{l}\mathbf{v}=\mathbf{\omega }\mathbf{X}\mathbf{r}\end{array}$

A close scrutiny of the quantities on the right hand of the expression of velocity indicate two possible changes :

• change in angular velocity ( ω ) and
• change in position vector ( r )

The angular velocity ( ω ) can change either in its direction (clockwise or anti-clockwise) or can change in its magnitude. There is no change in the direction of axis of rotation, however, which is fixed. As far as position vector ( r ) is concerned, there is no change in its magnitude i.e. | r | or r is constant, but its direction keeps changing with time. So there is only change of direction involved with vector “ r ”.

Now differentiating the vector equation, we have

$\begin{array}{l}\frac{đ\mathbf{v}}{đt}=\frac{đ\mathbf{\omega }}{đt}\mathbf{X}\mathbf{r}+\mathbf{\omega }\mathbf{X}\frac{đ\mathbf{r}}{đt}\end{array}$

We must understand the meaning of each of the acceleration defined by the differentials in the above equation :

• The term " $\frac{đ\mathbf{v}}{đt}$ " represents total acceleration ( a ) i.e. the resultant of radial ( ${\mathbf{a}}_{\mathbf{R}}$ ) and tangential acceleration( ${\mathbf{a}}_{\mathbf{T}}$ ).
• The term $\frac{đ\mathbf{\omega }}{đt}$ represents angular acceleration ( α )
• The term $\frac{đ\mathbf{r}}{đt}$ represents velocity of the particle ( v )

$\begin{array}{l}⇒\mathbf{a}=\mathbf{\alpha }\mathbf{X}\mathbf{r}+\mathbf{\omega }\mathbf{X}\mathbf{v}\end{array}$

$\begin{array}{l}⇒\mathbf{a}={\mathbf{a}}_{\mathbf{T}}+{\mathbf{a}}_{\mathbf{R}}\end{array}$

where, ${\mathbf{a}}_{\mathbf{T}}=\mathbf{\alpha }\mathbf{X}\mathbf{r}$ is tangential acceleration and is measure of the time rate change of the magnitude of the velocity of the particle in the tangential direction and ${\mathbf{a}}_{\mathbf{R}}=\mathbf{\omega }\mathbf{X}\mathbf{v}$ is the radial acceleration also known as centripetal acceleration, which is measure of time rate change of the velocity of the particle in radial direction.

Various vector quantities involved in the equation are shown graphically with respect to the plane of motion (xz plane) :

The magnitude of total acceleration in general circular motion is given by :

$\begin{array}{l}a=|\mathbf{a}|=\sqrt{\left({{a}_{T}}^{2}+{{a}_{R}}^{2}\right)}\end{array}$

Problem : At a particular instant, a particle is moving with a speed of 10 m/s on a circular path of radius 100 m. Its speed is increasing at the rate of 1 $m/{s}^{2}$ . What is the acceleration of the particle ?

Solution : The acceleration of the particle is the vector sum of mutually perpendicular radial and tangential accelerations. The magnitude of tangential acceleration is given here to be 1 $m/{s}^{2}$ . Now, the radial acceleration at the particular instant is :

$\begin{array}{l}{a}_{R}=\frac{{v}^{2}}{r}=\frac{{10}^{2}}{100}=1\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$

Hence, the magnitude of the acceleration of the particle is :

$\begin{array}{l}a=|\mathbf{a}|=\sqrt{\left({{a}_{T}}^{2}+{{a}_{R}}^{2}\right)}=\sqrt{\left({1}^{2}+{1}^{2}\right)}=\sqrt{2}\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$

Exercises

A particle is moving along a circle in yz - plane with varying linear speed. Then

(a) acceleration of the particle is in x – direction

(b) acceleration of the particle lies in xy – plane

(c) acceleration of the particle lies in xz – plane

(d) acceleration of the particle lies in yz – plane

The figure here shows the acceleration of the particle as the resultant of radial and tangential accelerations. The resultant acceleration lies in the plane of motion i.e yz – plane.

Hence, option (d) is correct.

A particle is moving along a circle of radius “r”. The linear and angular velocities at an instant during the motion are “v” and “ω” respectively. Then, the product vω represents :

(a) centripetal acceleration

(b) tangential acceleration

(c) angular acceleration divided by radius

(d) None of the above

The given product expands as :

$\begin{array}{l}v\omega =vx\frac{v}{r}=\frac{{v}^{2}}{r}\end{array}$

This is the expression of centripetal acceleration.

Hence, option (a) is correct.

Which of the following expression represents the magnitude of centripetal acceleration :

$\begin{array}{l}\mathrm{\left(a\right)}\phantom{\rule{2pt}{0ex}}|\frac{{đ}^{2}\mathbf{r}}{đ{t}^{2}}|\phantom{\rule{2pt}{0ex}}\mathrm{\left(b\right)}\phantom{\rule{2pt}{0ex}}|\frac{đ\mathbf{v}}{đt}|\phantom{\rule{2pt}{0ex}}\mathrm{\left(c\right)}\phantom{\rule{2pt}{0ex}}r\frac{đ\theta }{đt}\phantom{\rule{2pt}{0ex}}\mathrm{\left(d\right)}\phantom{\rule{2pt}{0ex}}\text{None of these}\end{array}$

The expression $|\frac{đ\mathbf{v}}{đt}|$ represents the magnitude of total or resultant acceleration. The differential $\frac{đ\theta }{đt}$ represents the magnitude of angular velocity. The expression $r\frac{đ\theta }{đt}$ represents the magnitude of tangential velocity. The expression $\frac{{đ}^{2}\mathbf{r}}{đ{t}^{2}}$ is second order differentiation of position vector ( r ). This is actually the expression of acceleration of a particle under motion. Hence, the expression $|\frac{{đ}^{2}\mathbf{r}}{đ{t}^{2}}|$ represents the magnitude of total or resultant acceleration.

Hence, option (d) is correct.

A particle is circling about origin in xy-plane with an angular speed of 0.2 rad/s. What is the linear speed (in m/s) of the particle at a point specified by the coordinate (3m,4m) ?

(a)1 (b) 2 (c) 3 (d) 4

The linear speed “v” is given by :

$\begin{array}{l}v=\omega r\end{array}$

Now radius of the circle is obtained from the position data. Here, x = 3 m and y = 4 m. Hence,

$\begin{array}{l}r=\sqrt{\left({3}^{2}+{4}^{2}\right)}=5\phantom{\rule{2pt}{0ex}}m\\ ⇒v=0.2x5=1\phantom{\rule{2pt}{0ex}}m/s\end{array}$

Hence, option (a) is correct.

A particle is executing circular motion. Man The velocity of the particle changes from zero to (0.3 i + 0.4 j ) m/s in a period of 1 second. The magnitude of average tangential acceleration is :

$\begin{array}{llllllll}\mathrm{\left(a\right)}& 0.1\phantom{\rule{2pt}{0ex}}m/{s}^{2}& \mathrm{\left(b\right)}& 0.2\phantom{\rule{2pt}{0ex}}m/{s}^{2}& \mathrm{\left(c\right)}& 0.3\phantom{\rule{2pt}{0ex}}m/{s}^{2}& \mathrm{\left(d\right)}& 0.5\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$

The magnitude of average tangential acceleration is the ratio of the change in speed and time :

$\begin{array}{l}{a}_{T}=\frac{\Delta v}{\Delta t}\end{array}$

Now,

$\begin{array}{l}\Delta v=\sqrt{\left({0.3}^{2}+{0.4}^{2}\right)}=\sqrt{0.25}=0.5\phantom{\rule{2pt}{0ex}}m/s\\ {a}_{T}=0.5\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$

Hence, option (d) is correct.

The radial and tangential accelerations of a particle in motions are ${a}_{T}$ and ${a}_{R}$ respectively. The motion can be circular if :

$\begin{array}{llllllll}\mathrm{\left(a\right)}\phantom{\rule{2pt}{0ex}}& {a}_{R}\ne 0,\phantom{\rule{2pt}{0ex}}{a}_{T}\ne 0& \phantom{\rule{2pt}{0ex}}\mathrm{\left(b\right)}\phantom{\rule{2pt}{0ex}}& {a}_{R}=0,\phantom{\rule{2pt}{0ex}}{a}_{T}=0& \phantom{\rule{2pt}{0ex}}\mathrm{\left(c\right)}\phantom{\rule{2pt}{0ex}}& {a}_{R}=0,\phantom{\rule{2pt}{0ex}}{a}_{T}=\ne & \phantom{\rule{2pt}{0ex}}\mathrm{\left(d\right)}\phantom{\rule{2pt}{0ex}}& {a}_{R}=\ne ,\phantom{\rule{2pt}{0ex}}{a}_{T}=0\end{array}$

Centripetal acceleration is a requirement for circular motion and as such it should be non-zero. On the other hand, tangential acceleration is zero for uniform circular acceleration and non-zero for non-uniform circular motion. Clearly, the motion can be circular motion if centripetal acceleration is non-zero.

Hence, options (a) and (d) are correct.

Which of the following pair of vector quantities is/are parallel to each other in direction ?

(a) angular velocity and linear velocity

(b) angular acceleration and tangential acceleration

(c) centripetal acceleration and tangential acceleration

(d) angular velocity and angular acceleration

The option (d) is correct.

A particle is moving along a circle in a plane with axis of rotation passing through the origin of circle. Which of the following pairs of vector quantities are perpendicular to each other :

(a) tangential acceleration and angular velocity

(b) centripetal acceleration and angular velocity

(c) position vector and angular velocity

(d) angular velocity and linear velocity

Clearly, vector attributes in each given pairs are perpendicular to each other.

Hence, options (a), (b), (c) and (d) are correct.

A particle is executing circular motion along a circle of diameter 2 m, with a tangential speed given by v = 2t.

(a) Tangential acceleration directly varies with time.

(b) Tangential acceleration inversely varies with time.

(c) Centripetal acceleration directly varies with time.

(d) Centripetal acceleration directly varies with square of time.

Tangential acceleration is found out by differentiating the expression of speed :

$\begin{array}{l}{a}_{T}=\frac{đv}{đt}\frac{đ\left(\mathrm{2t}\right)}{đt}=2\phantom{\rule{2pt}{0ex}}m/s\end{array}$

The tangential acceleration is a constant. Now, let us determine centripetal acceleration,

$\begin{array}{l}{a}_{R}=\frac{{v}_{2}}{r}=\frac{4{t}_{2}}{1}=4{t}_{2}\end{array}$

The option (d) is correct.

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