Calculating Torque in a Rolling Ball on an Inclined Track | Homework Problem

[myoplex11]

Homework Statement

A 4.74 kg spherical ball with radius 1.4 cm rolls on a track with a loop of radius 39 cm that sits on a table 1.2 meters above the ground. The last 1 meter section of the track is horzontal and is not connected to the rest of the track. 0.33 meters of this section of the track hangs over the edge of the table. The track has a linear mass density of 0.82 kg/m.
How far does the ball travel along the loose length of track before the track starts to tip?

Homework Equations

The Attempt at a Solution

can you please check my work thanks
Torqueball = Torquetrack
4.75(g)x = 0.82g(0.5)
x = .82/.4.75 (0.5) = 0.086498
it travels 0.33m - 0.089498 =0.24m

 

[tiny-tim]

Torqueball = Torquetrack
4.75(g)x = 0.82g(0.5)
x = .82/.4.75 (0.5) = 0.086498
it travels 0.33m - 0.089498 =0.24m

Hi myoplex11!

Nooo …

which point have you taken moments (torque) about?

take moments about the edge of the table …

try again!

 

[thomate1]

along the loose length of track before the track starts to tip.


I would like to provide a more detailed explanation and approach to solving this problem. Firstly, let's define the variables and equations that we will be using in this problem.

Variables:
m = mass of the ball (4.74 kg)
r = radius of the ball (0.014 m)
R = radius of the loop (0.39 m)
h = height of the table (1.2 m)
l = length of the loose section of the track (0.33 m)
μ = linear mass density of the track (0.82 kg/m)
g = acceleration due to gravity (9.8 m/s^2)
x = distance traveled along the loose length of track before tipping starts

Equations:
Torque = Force * Distance
Moment of Inertia = (2/5) * m * r^2
Gravitational Potential Energy = m * g * h
Kinetic Energy = (1/2) * m * v^2
Velocity of Rolling Ball = ω * R (where ω is the angular velocity)

Now, let's start by calculating the torque of the ball and the track. The torque of the ball is due to its weight and can be calculated as:
Tball = m * g * R * sinθ (where θ is the angle of inclination of the track)

The torque of the track can be calculated as:
Ttrack = (μ * l) * g * R * cosθ

Since the ball and the track are in equilibrium, the torque of the ball must be equal to the torque of the track. Therefore, we can equate the two equations and solve for θ:
m * g * R * sinθ = (μ * l) * g * R * cosθ
sinθ = (μ * l) * cosθ
tanθ = μ * l
θ = tan^-1(μ * l)

Now, using this value of θ, we can calculate the distance x using the equation:
x = R * (1 - cosθ)

Substituting the values of R and θ, we get:
x = (0.39 m) * (1 - cos(tan^-1(μ * l)))

Finally, substituting the values of μ and l, we get:
x = (0.39 m) * (1 -