Rock's Angular Motion in Bicycle Wheel Braking

You can convert between radians and degrees, but the numerical value will remain the same. So your answers are correct.
  • #1
freak_boy186
8
0

Homework Statement


A rock stuck in the tread of a 53.0 cm diameter bicycle wheel has a tangential speed of 3.50 m/s. When the brakes are applied, the rock's tangential deceleration is 1.20 m/s^2.

X.) What is the magnitudes of the rock's angular velocity (omega) when t = 1.60s?
Y.) What is the magnitudes of the rock's angular acceleration (alpha) when t = 1.60s?
Z.) At what time is the magnitude of the rock's acceleration equal to g? (9.8 m/s^2)

Homework Equations


v = r(omega)
a = r(omega)^2
Vf = Vi + a[tangental](dt)
D[omega] = alpha(dt)
omega^2 = omega0^2 + 2(alpha)(dtheta)

The Attempt at a Solution


X.) r = 0.265m, a[tangental] = -1.2, Vi = 3.5, dt = 1.6
Vf = 3.5 + (-1.2)(1.6) = 1.58
omega = 1.58/0.265 = 5.962 {incorrect apparently}

Y.) omega1 = 3.5/.265 = 13.208
(13.208 - 5.962) = alpha(1.6)
alpha = 4.528 {incorrect, but expected because X is incorrect}
 
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  • #2
freak_boy186 said:

Homework Statement


A rock stuck in the tread of a 53.0 cm diameter bicycle wheel has a tangential speed of 3.50 m/s. When the brakes are applied, the rock's tangential deceleration is 1.20 m/s^2.

X.) What is the magnitudes of the rock's angular velocity (omega) when t = 1.60s?
Y.) What is the magnitudes of the rock's angular acceleration (alpha) when t = 1.60s?
Z.) At what time is the magnitude of the rock's acceleration equal to g? (9.8 m/s^2)

Homework Equations


v = r(omega)
a = r(omega)^2
Vf = Vi + a[tangental](dt)
D[omega] = alpha(dt)
omega^2 = omega0^2 + 2(alpha)(dtheta)

The Attempt at a Solution


X.) r = 0.265m, a[tangental] = -1.2, Vi = 3.5, dt = 1.6
Vf = 3.5 + (-1.2)(1.6) = 1.58
omega = 1.58/0.265 = 5.962 {incorrect apparently}
That's correct. Maybe you're not entering it in the right units or with the correct number of significant figures.
Y.) omega1 = 3.5/.265 = 13.208
(13.208 - 5.962) = alpha(1.6)
alpha = 4.528 {incorrect, but expected because X is incorrect}
That's correct too, although you took a roundabout way of calculating it.
 
  • #3
its asking for omega in rad/sec & alpha in rad/sec^2... would that affect my answers any?
 
  • #4
Nope. Radians is the natural unit for measuring angles.
 

Related to Rock's Angular Motion in Bicycle Wheel Braking

1. What is circular motion?

Circular motion is the movement of an object along a circular path, where the object's distance from a fixed point remains constant while its direction continuously changes.

2. How does a bicycle move in a circular motion?

A bicycle moves in a circular motion by the rotation of its wheels, which are attached to the pedals. When the pedals are pushed, the wheels rotate, causing the bicycle to move forward in a circular path.

3. What causes the centripetal force in circular motion of a bicycle?

The centripetal force in circular motion of a bicycle is caused by the inward pull of the tension in the spokes of the wheels. This force is necessary to keep the bicycle moving in a circular path.

4. How does the radius of the circle affect the speed of a bicycle in circular motion?

The speed of a bicycle in circular motion is affected by the radius of the circle it is moving in. A smaller radius will result in a higher speed, while a larger radius will result in a slower speed. This is because a smaller radius requires the bicycle to travel a shorter distance in the same amount of time.

5. What is the relationship between the centripetal force and the speed of a bicycle in circular motion?

The centripetal force and the speed of a bicycle in circular motion are directly proportional. This means that as the speed of the bicycle increases, the centripetal force also increases in order to maintain the circular motion. Similarly, if the speed decreases, the centripetal force decreases as well.

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