Ball sliding on 3d-inclined plane

AI Thread Summary
The discussion revolves around determining the time it takes for a solid ball to roll down an inclined plane without slipping. Two methods were used to solve the problem: one based on rotational dynamics and linear kinematics, and the other on conservation of energy. The first method yielded a time expression of t = √(14L/5g sinΘ), while the second produced t = √(10L/7g sin Θ). The discrepancy in results was resolved by recognizing that the acceleration should account for friction, leading to the conclusion that the first method's assumption of a = g sin Θ was incorrect. The problem was ultimately solved by correcting the acceleration used in the kinematic equations.
terryds
Messages
392
Reaction score
13

Homework Statement



http://www.sumoware.com/images/temp/xzdmdjtlksnfhiqc.png
A solid ball rolls perfectly with initial velocity v0 in horizontal axis ( y-axis ) on an inclined plane with elevation angle Θ as the picture above shown. This ball moves turning due to the gravitational acceleration till it has traveled distance L in x-axis when it's at the bottom of the plane.
Determine the time (t) the ball needs to get to the bottom of the plane ! (The ball doesn't slip while rolling)

Homework Equations


Rotational dynamics equation and linear kinematics equation
Or conservation of energy equation

The Attempt at a Solution



I have two methods to solve the problem. But, the answers are different.

Using rot. dynamics equation and linear kinematics.
I just consider the x-axis since it's what the question asks.
ΣFx = ma
mg sin Θ - f = ma (Note : f is friction force)
f = mg sin Θ - ma

Στ = I α
f R = I α
f R = I (a/R)
(mg sin Θ - ma) R = (2/5) m R^2 (a/R)
mg sin Θ - ma = (2/5) m a
g sin Θ - a = (2/5) a
(7/5) a = g sin Θ
a = (5/7) g sin Θ

Then, I use the kinematics equation
L = 0.5 a t^2
2L/a = t^2
14L/ (5g sin Θ) = t^2
t = √(14L/5g sinΘ)

But, using conservation of energy, I get different answer
m g sin Θ L = (1/2) m v^2 + (1/2) I ω^2
m g sin Θ L = (1/2) m v^2 + (1/2) (2/5 m R^2) (v^2 / R^2)
g sin Θ L = (1/2) v^2 + (1/5) m v^2
g sin Θ L = (7/10) v^2
v = √(10 g sin Θ L / 7 )

Then, I use the kinematics
vt = vox + a t
√(10 g sin Θ L / 7 ) = 0 + g sin Θ t
t = √(10L/7g sin Θ)


Which one is correct? Why?
 
Last edited by a moderator:
Physics news on Phys.org
terryds said:

Homework Statement



http://www.sumoware.com/images/temp/xzdmdjtlksnfhiqc.png
A solid ball rolls perfectly with initial velocity v0 in horizontal axis ( y-axis ) on an inclined plane with elevation angle Θ as the picture above shown. This ball moves turning due to the gravitational acceleration till it has traveled distance L in x-axis when it's at the bottom of the plane.
Determine the time (t) the ball needs to get to the bottom of the plane ! (The ball doesn't slip while rolling)

Homework Equations


Rotational dynamics equation and linear kinematics equation
Or conservation of energy equation

The Attempt at a Solution



I have two methods to solve the problem. But, the answers are different.

Using rot. dynamics equation and linear kinematics.
I just consider the x-axis since it's what the question asks.
ΣFx = ma
mg sin Θ - f = ma (Note : f is friction force)
f = mg sin Θ - ma

Στ = I α
f R = I α
f R = I (a/R)
(mg sin Θ - ma) R = (2/5) m R^2 (a/R)
mg sin Θ - ma = (2/5) m a
g sin Θ - a = (2/5) a
(7/5) a = g sin Θ
a = (5/7) g sin Θ

Then, I use the kinematics equation
L = 0.5 a t^2
2L/a = t^2
14L/ (5g sin Θ) = t^2
t = √(14L/5g sinΘ)

But, using conservation of energy, I get different answer
m g sin Θ L = (1/2) m v^2 + (1/2) I ω^2
m g sin Θ L = (1/2) m v^2 + (1/2) (2/5 m R^2) (v^2 / R^2)
g sin Θ L = (1/2) v^2 + (1/5) m v^2
g sin Θ L = (7/10) v^2
v = √(10 g sin Θ L / 7 )

Then, I use the kinematics
vt = vox + a t
√(10 g sin Θ L / 7 ) = 0 + g sin Θ t
t = √(10L/7g sin Θ)


Which one is correct? Why?

Oops.. I think that I forgot that the acceleration is not g sin theta since there is friction.. ( I used a = g sin theta in using the kinematics )
Problem Solved :)
 
Last edited by a moderator:
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 '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