Question

Starting from rest, Ali goes around a circular path (of 100 m diameter) with a constant
angular acceleration a. If he completed 4 rounds every 3 min, what is his angular
displacement, and angular acceleration and final angular speed, ?

More information is needed
8, 0.56 rad/sec2, 101 rad/sec
800, 0.078 rad/sec, 12.4 rad/sec2
8 , 0.003 rad/sec2, 0.56 rad/sec
800, 0.078 rad/sec2, 12.4 rad/sec

Answer 1

Answer on Question #41002, Physics, Mechanics

Starting from rest, Ali goes around a circular path (of $100\mathrm{m}$ diameter) with a constant angular acceleration $\alpha$ . If he completed 4 rounds every $3\mathrm{min}$ , what is his angular displacement, and angular acceleration and final angular speed?

More information is needed

8, 0.56 rad/sec², 101 rad/sec

800, 0.078 rad/sec, 12.4 rad/sec²

8, 0.003 rad/sec², 0.56 rad/sec

800, 0.078 rad/sec², 12.4 rad/sec

Solution:

Given:

$d = 100\mathrm{m}$ radius is $r = 50\mathrm{m}$

$t = 3\mathrm{min} = 180\mathrm{s}$

Time for one round is $t_1 = \frac{180}{4} s = 45$ s,

$\omega_0 = 0$

$\theta = ?$

$\alpha = ?$

$\omega = ?$

The angular displacement is defined by:

$$\theta = \frac {S}{r} = \frac {4 \cdot 2 \pi r}{r} = 8 \pi = 2 5. 1 2 \mathrm {r a d}$$

Equations for constant angular acceleration

$$\theta = \overline {{\omega}} t, \overline {{\omega}} = \frac {\omega_ {0} + \omega}{2} = \frac {0 + \omega}{2} = \frac {\omega}{2}$$ $$\omega^ {2} = \omega_ {0} ^ {2} + 2 \alpha \theta$$

Thus, the final angular speed is

$$\omega = \frac {2 \theta}{t} = \frac {2 \cdot 2 5 . 1 2}{1 8 0} = 0. 2 8 \mathrm {r a d / s}$$ $$\theta = \omega_ {0} t + \frac {1}{2} \alpha t ^ {2}$$

The angular acceleration is

$$\alpha = \frac {\omega^ {2}}{2 \theta} = \frac {0 . 2 8 ^ {2}}{2 \cdot 2 5 . 1 2} = 0. 0 0 1 5 6 \mathrm {r a d / s ^ {2}}$$

Answer. More information is needed.

The angular displacement is $\theta = 25.12$ rad;

the angular acceleration is $\alpha = 0.00156\mathrm{rad / s^2}$

the final angular speed is $\omega = 0.28$ rad/s.

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