Question

You set a small eraser at a distance of 0.27 m from the center of a turntable you find in the attic and establish that when the turntable is turned on, the eraser travels in a circle with a constant tangential speed of 0.49 m/s. If your desk lamp projects a shadow of the apparatus on a wall which is the same size as the real apparatus, determine the following for the oscillatory motion of the eraser's shadow.
(a) period
s

(b) amplitude
m

(c) maximum speed
m/s

(d) magnitude of the maximum acceleration
m/s2

Answer 1

Answer with Explanation:

We are given that

Distance,r=0.27 m

Tangential speed=v=0.49 m/s

a.Angular speed ,$\omega=\frac{v}{r}$

Using the formula

$\omega=\frac{0.49}{0.27}=1.8 rad/s$

$\omega=\frac{2\pi}{T}$

$T=\frac{2\pi}{\omega}$

Time period,$T=\frac{2\pi}{1.8}=3.49 s$

b.Amplitude,A=Distance of small eraser from the center of a turnable =0.27 m

c.Maximum speed,$V_{max}=0.49 m/s$

d.Maximum acceleration=$a=r\omega^2=0.27(1.8)^2=0.87 m/s^2$

Answer 2

Answer:

Explanation:

radius of path, r = 0.27 m

tangential velocity, v = 0.49 m/s

(a) Let the period is T.

Angular speed, ω = v/r

ω = 0.49 / 0.27

ω = 1.81 rad/s

T = 2π/ω

T = ( 2 x 3.14) / 1.81

T = 3.47 s

(b) Amplitude, A = r = 0.27 m

(c) maximum velocity, v = ωA = 1.81 x 0.27 = 0.49 m/s

(d maximum acceleration, a = ω²A = 1.81 x 1.81 x 0.27 = 0.885 m/s²