Answer to Question #280769 in Physics for tetsuya

Question

1. A wheel started from rest and attained an angular speed of 200 rev/s in 1 minute. Calculate

its angular acceleration assuming it to be constant.

2. A rotating body makes 100 revolutions in 10 seconds. Finds its angular speed.

3. A wheel of radius 0.25 m that starts from rest is given an angular acceleration of 10 rad/s

2

.

Find:

(a) the at of point at its rim,

(b) the ac of this point after 5 seconds, and

(c) the actual acceleration of the wheel.

4. Find the total torque acting on a body when five forces are applied on the following

moment-arms: F1 = 30 N upward at r1 = 2.0 m, F2 = 10 N upward at r2 = 3.0 m, F3 = 50 N

downward at r3 = 2.0 m, F4 = 20 N upward at r4 = 0.5 m and F5 = 25 N downward at r5 = 1.5 m.

Answer 1

1)

$$\alpha=\dfrac{\omega-\omega_0}{t}=\dfrac{200\ \dfrac{rev}{s}\times\dfrac{2\pi\ rad}{1\ rev}-0}{60\ s}=20.94\ \dfrac{rad}{s^2}.$$

2) Let's first find the angular displacement covered during 100 revolutions:

$$\theta=\dfrac{2\pi\ rad}{1\ rev}\times n=\dfrac{2\pi\ rad}{1\ rev}\times 100\ rev=628\ rad.$$

Then, we can find the angular acceleration of the body from the kinematic equation:

$$\theta=\omega_0t+\dfrac{1}{2}\alpha t^2,$$ $$\theta=\dfrac{1}{2}\alpha t^2,$$ $$\alpha=\dfrac{2\theta}{t^2}=\dfrac{2\times628\ rad}{(10\ s)^2}=12.56\ \dfrac{rad}{s^2}.$$

Finally, we can find the angular speed of the body as follows:

$$\omega=\omega_0+\alpha t=0+12.56\ \dfrac{rad}{s^2}\times10\ s=125.6\ \dfrac{rad}{s}.$$

3) (a)

$$a_t=\alpha r=0.25\ m\times10\ \dfrac{rad}{s^2}=2.5\ \dfrac{m}{s^2}.$$

b)

$$\omega=\omega_0+\alpha t=0+10\ \dfrac{rad}{s^2}\times5\ s=50\ \dfrac{rad}{s},$$ $$a_c=\dfrac{v^2}{r}=\omega^2r=(50\ \dfrac{rad}{s})^2\times0.25\ m=625\ \dfrac{m}{s^2}.$$

(c)

$$a=\sqrt{a_t^2+a_c^2}=\sqrt{(2.5\ \dfrac{m}{s^2})^2+(625\ \dfrac{m}{s^2})^2}=625\ \dfrac{rad}{s^2}.$$

4)

$$F_{tot}=F_1r_1+F_2r_2-F_3r_3+F_4r_4-F_5r_5,$$

$F_{tot}=30\ N\times2\ m+10\ N\times3\ m-50\ N\times2\ m+20\ N\times0.5\ m-25\ N\times1.5\ m,$

$$F_{tot}=-37.5\ N\times m.$$

The sign minus means that the total torque acting on the body directed downward.