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

Step-by-step 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

Answer 2

Answer with Explanation:

We are given that

Distance,r=0.27 m

Tangential speed=v=0.49 m/s

a.Angular speed ,
`\omega=(v)/(r)`

Using the formula

`\omega=(0.49)/(0.27)=1.8 rad/s`

`\omega=(2\pi)/(T)`

`T=(2\pi)/(\omega)`

Time period,
`T=(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`