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 \[{{\omega }_{2}}={{\omega }_{1}}-\alpha \,t\]\[\Rightarrow \]\[0={{\omega }_{1}}-\alpha \,t\] \[t=\frac{{{\omega }_{1}}}{\alpha }=\frac{10\times 6}{5}=12\,sec\].

Notes - Equations of Circular Motion
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