# Non-uniform Circular Motion

### Uniform Angular Acceleration  ?

If the magnitude of velocity of the revolving particle in a circular path changes uniformly, we say that the particle is accelerating tangentially.

Let the tangential acceleration at any time be , the corresponding instantaneous velocities (as shown in the Figure) at 1 and 2 are $\displaystyle \vec{v_1} \; and \; \vec{v_2}$ respectively and the respective instantaneous angular speeds at 1 and 2 be ω1 and ω2 respectively.

Since $\displaystyle \omega = \frac{v}{r} \; and \; v = \frac{v_1 + v_2}{2}$

the average angular velocity during time t can be given as

$\displaystyle \omega = \frac{\omega_1 + \omega_2}{2}$

and the angular acceleration , $\displaystyle \alpha = \frac{\Delta \omega}{\Delta t}$

$\displaystyle \alpha = \frac{\omega_2 -\omega_1}{t}$

$\displaystyle \omega_2 = \omega_1 + \alpha t$

Putting , ω1 = ω0 , ω2 = ω and Δt = t

ω = ω0 + α t

Since the angular distance θ covered during time t is given as

$\displaystyle \theta = \omega t = \frac{\omega_1 + \omega_2}{2} t$

$\displaystyle \theta = \frac{\omega_0 + \omega}{2} t$

Putting , $\displaystyle t = \frac{\omega – \omega_0}{\alpha}$

We obtain

$\displaystyle \theta = \frac{\omega^2 – \omega_0^2}{2 \alpha}$

$\displaystyle \omega^2 = \omega_0^2 + 2 \alpha \theta$

Putting , $\displaystyle \omega = \omega_0 + \alpha t$

We obtain

$\displaystyle \theta = \omega_0 t + \frac{1}{2}\alpha t^2$

Since , ω = v/r ; Δω = Δv/r

$\displaystyle \Delta \omega/\Delta t = \frac{\Delta v / \Delta t}{r}$

When Δt → 0

$\displaystyle \frac{d\omega}{dt} = \frac{1}{r}\frac{dv}{dt}$

Since , $\displaystyle \frac{dv}{dt} = a_t \; , and \; \frac{d\omega}{dt} = \alpha$

$\displaystyle \alpha = \frac{1}{r}a_t$

$\displaystyle a_t = r \alpha$

Solved Example : A particle of mass (m) is moving in a circular path of constant radius with centripetal acceleration , ac = K2rt2 find power delivered to the particle by the force acting on it .

Sol: Since,  $\displaystyle \frac{v^2}{r} = K^2 r t^2$

$\displaystyle v = K r t$

Differentiating w.r.t time

$\displaystyle \frac{dv}{dt} = K r$

a t = K r

F = m a t = m K r

P = F .v

P = m K r (K r t)

P = m K2 r2 t

Solved Example : Particle moves on a circular path of radius (r). Its kinetic energy is related with distance such as , K = a s2, where a = const. Find out the total force on the particle in terms of distance (s) .

Sol: $\displaystyle K = \frac{1}{2} m v^2 = a s^2$

$\displaystyle v^2 = \frac{2 a s^2}{m}$

$\displaystyle F_c = \frac{m v^2}{r}$

$\displaystyle F_c = \frac{m}{r}.\frac{2 a s^2}{m}= \frac{2 a s^2}{r}$

$\displaystyle F_t = m . a_t = m \frac{dv}{dt}$

Since , $\displaystyle v^2 = \frac{2 a s^2}{m}$

$\displaystyle 2 v \frac{dv}{dt} = \frac{4 a s}{m} \frac{ds}{dt}$

$\displaystyle v \frac{dv}{dt} = \frac{2 a s}{m} v$

$\displaystyle a_t = \frac{dv}{dt} = \frac{2 a s}{m}$

$\displaystyle F_t = m . a_t = m \frac{2 a s}{m} = 2 a s$

$\displaystyle F = \sqrt{F_t ^2 + F_c ^2}$

$\displaystyle F = \sqrt{4 a s^2 + \frac{4 a^2 s^4}{r^2}}$

$\displaystyle F = 2 a s \sqrt{1 + \frac{s^2}{r^2}}$

Exercise : Referring to the previous illustration, find the (a) average angular speed of the disc (b) total number of revolutions ‘N’ of the disc before coming to rest.

### Non-uniform Angular Acceleration ?

If ω and  α are not constant in magnitude, that means α = f (t) or α = f(θ) we write the kinematics  equations as

$\displaystyle \theta = \int_{0}^{t} \omega dt$

$\displaystyle \int_{\omega_0}^{\omega} \omega d\omega = \int_{0}^{\theta} \alpha d\theta$

$\displaystyle \omega^2 = \omega_0^2 + 2\int_{0}^{\theta} \alpha d\theta$

Q: A particle of mass m is moving on a circular path of radius r in the influence of force $F = \frac{-k}{r^2} \hat{r}$ , where k is a positive constant.

(a) Kinetic energy of particle is k/r

(b) Kinetic energy of particle k/2r

(c) Potential energy of particle is (-k)/(r )

(d) Total mechanical energy of particle is (– k )/2r

Click to See Answer :
Ans: (b) , (c) , (d)

Q: A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near to top of the wire and it slides along the wire without friction. As, the bead moves from A to B, the force it applies on the wire is

(a) Always radially outwards

(b) Always radially inwards

(c) Radially outwards initially and radially inwards later

(d) Radially inwards initially and radially outwards later

Click to See Answer :
Ans: (d)

Q: A cyclist riding at a speed of 9.8 m/s takes a turn around a circular road of radius 19.6 m. What is his inclination to the vertical?

(a) 30°

(b) 45°

(c) 10.5°

(d) 26.5°

Click to See Answer :
Ans: (d)