Question

An exhausted bicyclist pedals somewhat erratically when exercising on a static bicycle. The angular velocity of the wheels follows the equation ?(t)=at?bsin(ct)fort?0, where t represents time (measured in seconds), a = 0.500 rad/s2 , b = 0.250 rad/s and c = 2.00 rad/s . A There is a spot of paint on the front wheel of the bicycle. Take the position of the spot at time t=0 to be at angle ?=0 radians with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the wheel's rotation. What angular displacement ? has the spot of paint undergone between time 0 and 2 seconds? BExpress the angular displacement undergone by the spot of paint at t=2 seconds in degrees. Remember to use the unrounded value from Part A, should you need it. CWhat distance d has the spot of paint moved in 2 seconds if the radius of the wheel is 50 centimeters?

Answer 1

Answer:

A: Θ=.793 radians

B: Θ= 45.5 degrees

C: d= 39.7 centimeters

D: The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is increasing.

Explanation:

Answer 2

The motion of the spot on the tire in a circular path can be expressed both as angular motion and translational motion

a) The angular displacement of the spot in the first 2 seconds is approximately 0.79 radians

b) The angular displacement in the first two seconds in degrees is approximately 45.452°

c) The distance the spot of paint moved in the first 2 seconds is approximately 39.5 centimeters

The reason by which the above values are arrived at are as follows:

The known parameters are;

The equation of the angular velocity of the wheel, is

ω(t) = a·t - b·sin(c·t) for t ≥ 0

Where;

t = The time measured in seconds

a = 0.500 rad/s²

b = 0.250 rad/s

c = 2.00 rad/s

The angular position of the spot on the wheel at time, t = 0,  θ = 0 radians w.r.t. the axis parallel to the ground  (horizontal axis) as well as perpendicular to the wheel's axis of rotation

The required parameter;

a) The angular displacement of the spot between t = 0, and t = 2 seconds

Solution:

Plugging in the known values gives;

ω(t) = 0.5 × t - 0.250 × sin(2.00 × t)

$Angular \ velocity, \ \omega (t)= \dfrac{d \theta}{dt}$

Therefore;

dθ = ω(t)·d(t)

$\displaystyle \int\limits d(\theta) = \theta = \int\limits {\omega(t)} \, dt$

$\displaystyle \theta = \int\limits_0^2 {\left(0.5 \times t - 0.250 \times sin(2.00 \times t)} \right) \, dt = \left 0.25 \cdot t^2 - 0.25 \cdot sin^2(t) \right | _0 ^2$

Therefore;

θ = (0.25×2² - 0.25×sin²(2)) - (0.25×0² - 0.25×sin²(0)) ≈ 0.79

The angular displacement of the spot between time, t = 0 seconds and time t = 2 seconds ≈ 0.79 radians

b) The measure of the angular displacement in degrees

The angular displacement undergone by the spot at t = 2 seconds in degrees is given as follows;

θ (Degrees) = 360/(2×π) (0.25×2² - 0.25×sin²(2)) - (0.25×0² - 0.25×sin²(0)) ≈ 45.452°

θ (Degrees) ≈ 45.452°

c) The translational distance moved by the spot

Given that the radius of the wheel, r = 50 centimeters, we have;

The (translational) distance moved by the spot, s = Arc length =  r × θ

∴ s ≈ 50 cm × 0.79 radians = 39.5 cm

The (translational) distance moved by the spot, s ≈ 39.5 cm

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