# NEET Physics Two Dimensional Motion Equations of Circular Motion

Equations of Circular Motion

Category : NEET

Equations of Circular Motion

 For accelerated motion For retarded motion ${{\omega }_{2}}={{\omega }_{1}}+\alpha \,t$ ${{\omega }_{2}}={{\omega }_{1}}-\alpha \,t$ $\theta ={{\omega }_{1}}t+\frac{1}{2}\alpha \,{{t}^{2}}$ $\theta ={{\omega }_{1}}t-\frac{1}{2}\alpha \,{{t}^{2}}$ $\omega _{_{2}}^{2}=\omega _{_{1}}^{2}+2\alpha \,\theta$ $\omega _{_{2}}^{2}=\omega _{_{1}}^{2}-2\alpha \,\theta$ ${{\theta }_{n}}={{\omega }_{1}}+\frac{\alpha }{2}(2n-1)$ ${{\theta }_{n}}={{\omega }_{1}}-\frac{\alpha }{2}(2n-1)$
Sample problems based on equation of circular motion Where   ${{\omega }_{1}}= Initial angular velocity of particle$ ${{\omega }_{2}}= Final angular velocity of particle$ $\alpha ~=\text{ }Angular\text{ }acceleration\text{ }of\text{ }particle$ $\theta = Angle covered by the particle in time t$ ${{\theta }_{n}}= Angle covered by the particle in {{n}^{th}}second$ Problem 150. The angular velocity of a particle is given by $\omega =1.5\ t-3{{t}^{2}}+2$, the time when its angular acceleration ceases to be zero will be (a) $25\ \sec$    (b) $0.25\ \sec$ (c) $12\ \sec$    (d) $1.2\ \sec$ Solution: (b) $\omega =1.5\,t-3{{t}^{2}}+2$ and $\alpha =\frac{d\omega }{dt}=1.5-6t$ $\Rightarrow \,\,0=1.5-6t$ $\therefore \,\,t=\frac{1.5}{6}=0.25\,\sec$ Problem 151. A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first $2\sec$, it rotates through an angle ${{\theta }_{1}}$. In the next $2\sec$, it rotates through an additional angle ${{\theta }_{2}}$. The ratio of  ${{\theta }_{1}}/{{\theta }_{2}}$ is [AIIMS 1982] (a) 1      (b) 2      (c) 3      (d) 5 Solution: (c) From equation of motion $\theta ={{\omega }_{1}}t+\frac{1}{2}\alpha \,{{t}^{2}}$ ${{\theta }_{1}}=0+\frac{1}{2}\alpha \,{{(2)}^{2}}=2\alpha$            ?.. (i)   [As ${{\omega }_{1}}=0,$ $t=2\,\sec ,$ $\theta ={{\theta }_{1}}$] For second condition ${{\theta }_{1}}+{{\theta }_{2}}=0+\frac{1}{2}\alpha \,{{(4)}^{2}}$             [As ${{\omega }_{1}}=0,$ $t=2+2=4\,\sec ,$ $\theta ={{\theta }_{1}}+{{\theta }_{2}}$] ${{\theta }_{1}}+{{\theta }_{2}}=8\alpha$     ?. (ii) From (i) and (ii) ${{\theta }_{1}}=2\alpha ,$${{\theta }_{2}}=6\alpha$$\therefore \,\,\frac{{{\theta }_{2}}}{{{\theta }_{1}}}=3$ Problem 152.   If the equation for the displacement of a particle moving on a circular path is given by$(\theta )=2{{t}^{3}}+0.5$, where $\theta$ is in radians and $t$in seconds, then the angular velocity of the particle after $2\sec$ from its start is [AIIMS 1998] (a) $8\ rad/\sec$            (b) $12\ rad/\sec$          (c) $24\ rad/\sec$          (d) $36\ rad/\sec$ Solution: (c) $\theta =2{{t}^{3}}+0.5$  and  $\omega =\frac{d\theta }{dt}=6{{t}^{2}}$ at t = 2 sec, $\omega =6\,{{(2)}^{2}}=24\,rad/sec$ Problem 153. A grinding wheel attained a velocity of 20 rad/sec in 5 sec starting from rest. Find the number of revolutions made by the wheel (a) $\frac{\pi }{25}$ rev/ sec       (b) $\frac{1}{\pi }$ rev/sec         (c) $\frac{25}{\pi }$ rev/sec        (d) None of these Solution: (c) ${{\omega }_{1}}=0,$ ${{\omega }_{2}}=20\,rad/sec,$ $t=5\,sec$ $\alpha =\frac{{{\omega }_{2}}-{{\omega }_{1}}}{t}=\frac{20-0}{5}=4\,rad/se{{c}^{2}}$ From the equation $\theta ={{\omega }_{1}}t+\frac{1}{2}\alpha \,{{t}^{2}}=0+\frac{1}{2}(4)\,.\,{{(5)}^{2}}$$=50\,rad$ $2\pi \,rad$ means 1 revolution.  $\therefore$ 50 Radian means $\frac{50}{2\pi }\,$or $\frac{25}{\pi }rev$. Problem 154. A grind stone starts from rest and has a constant angular acceleration of 4.0 rad/sec2. The angular displacement and angular velocity, after 4 sec. will respectively be (a) 32 rad, 16 rad/sec      (b) 16 rad, 32 rad/sec      (c) 64 rad, 32 rad/sec      (d) 32 rad, 64 rad/sec Solution: (a) ${{\omega }_{1}}=0,$$\alpha =4\,rad/{{\sec }^{2}},$$t=4\,sec$ Angular displacement $\theta ={{\omega }_{1}}t+\frac{1}{2}\alpha \,{{t}^{2}}$$=0+\frac{1}{2}4\,{{(4)}^{2}}$$=32\,rad.$ \ Final angular ${{\omega }_{2}}={{\omega }_{1}}+\alpha \,t=0+4\times 4=16\,rad/sec$ Problem 155.   An electric fan is rotating at a speed of 600 rev/minute. When the power supply is stopped, it stops after 60 revolutions. The time taken to stop is         (a) 12 s              (b) 30 s              (c) 45 s              (d) 60 s Solution: (a) ${{\omega }_{1}}=600\,\,\,rev/\min =10\,rev/sec,$ ${{\omega }_{2}}=0\,\,\text{and}\,\,\theta =60\,\,rev$ From the equation $\omega _{2}^{2}=\omega _{1}^{2}-2\alpha \theta$$\Rightarrow$$0={{(10)}^{2}}-2\,\alpha \,60$ $\therefore \alpha =\frac{100}{120}=\frac{5}{6}$ Again ${{\omega }_{2}}={{\omega }_{1}}-\alpha \,t$$\Rightarrow$$0={{\omega }_{1}}-\alpha \,t$ $t=\frac{{{\omega }_{1}}}{\alpha }=\frac{10\times 6}{5}=12\,sec$.

##### 30 20

You need to login to perform this action.
You will be redirected in 3 sec