What is the centripetal acceleration of a point on the perimeter of a bicycle wh

Question

What is the centripetal acceleration of a point on the perimeter of a bicycle wheel of diameter 70.0 cm when the bike is moving 8.00 m/s? It a guitar string has a fundamental frequency of 500 Hz, which one of the following frequencies can set the string into resonant vibration? A solid object floats in water with three-fourths of its volume beneath the surface. What is the object's density? A wheel of diameter of 68.0 cm slows down uniformly from 8.40 m/s to rest over a distance of 115 m. What is the angular acceleration?

Explanation / Answer

1. a_c = v^2 / R

R = 70/2 = 35 cm = 0.35 m

a_c = 8^2 / 0.35 =182.9 m/s^2

Ans(B)

2. for guitar,

f = v (n + 1) / 2L

fundamental f when n = 0

v/2L = 500 Hz

when n = 1

f = 500(1 + 1) = 1000 Hz

n = 2

f = 500 (2 + 1) = 1500 Hz

3. as object float that means it is in equilibrium.

Fnet = mg - Fb =0

Fb = buoyant force = weight of displaced water

(rho * V * g) - ( 1000 * (3V/4) * g) = 0

rho = 750 kg / m^3

4. applying ,

vf^2 - vi^2 = 2 a d

0^2 - 8.40^2 = 2 ( a) (115)

a = - 0.307 m/s^2

r = 68 cm /2 = 34 cm = 0.34 m

alpha = a/r = - 0.307 / 0.34 = - 0.903 rad/s^2

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