Rock's Angular Motion in Bicycle Wheel Braking

AI Thread Summary
A rock embedded in a bicycle wheel with a diameter of 53.0 cm has a tangential speed of 3.50 m/s and experiences a tangential deceleration of 1.20 m/s² when brakes are applied. The calculations for angular velocity (omega) and angular acceleration (alpha) are discussed, with initial attempts yielding incorrect results due to unit or significant figure errors. The correct approach involves using the relationship between tangential speed and radius to find omega and applying the angular acceleration formula. The conversation emphasizes the importance of using radians for angular measurements. Overall, the thread explores the physics of angular motion in the context of braking dynamics.
freak_boy186
Messages
8
Reaction score
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}
 
Physics news on Phys.org
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.
 
its asking for omega in rad/sec & alpha in rad/sec^2... would that affect my answers any?
 
Nope. Radians is the natural unit for measuring angles.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top