Question #280769

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.

1
Expert's answer
2021-12-20T10:27:56-0500

1)

α=ωω0t=200 revs×2π rad1 rev060 s=20.94 rads2.\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:


θ=2π rad1 rev×n=2π rad1 rev×100 rev=628 rad.\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:


θ=ω0t+12αt2,\theta=\omega_0t+\dfrac{1}{2}\alpha t^2,θ=12αt2,\theta=\dfrac{1}{2}\alpha t^2,α=2θt2=2×628 rad(10 s)2=12.56 rads2.\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:


ω=ω0+αt=0+12.56 rads2×10 s=125.6 rads.\omega=\omega_0+\alpha t=0+12.56\ \dfrac{rad}{s^2}\times10\ s=125.6\ \dfrac{rad}{s}.

3) (a)

at=αr=0.25 m×10 rads2=2.5 ms2.a_t=\alpha r=0.25\ m\times10\ \dfrac{rad}{s^2}=2.5\ \dfrac{m}{s^2}.

b)

ω=ω0+αt=0+10 rads2×5 s=50 rads,\omega=\omega_0+\alpha t=0+10\ \dfrac{rad}{s^2}\times5\ s=50\ \dfrac{rad}{s},ac=v2r=ω2r=(50 rads)2×0.25 m=625 ms2.a_c=\dfrac{v^2}{r}=\omega^2r=(50\ \dfrac{rad}{s})^2\times0.25\ m=625\ \dfrac{m}{s^2}.

(c)

a=at2+ac2=(2.5 ms2)2+(625 ms2)2=625 rads2.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)

Ftot=F1r1+F2r2F3r3+F4r4F5r5,F_{tot}=F_1r_1+F_2r_2-F_3r_3+F_4r_4-F_5r_5,

Ftot=30 N×2 m+10 N×3 m50 N×2 m+20 N×0.5 m25 N×1.5 m,F_{tot}=30\ N\times2\ m+10\ N\times3\ m-50\ N\times2\ m+20\ N\times0.5\ m-25\ N\times1.5\ m,

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

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS