What is the drum's angular acceleration?

AI Thread Summary
The angular acceleration of a drum rolling down a slope without slipping is given by the formula α = a/R, where 'a' is the linear acceleration parallel to the slope and 'R' is the radius of the drum. The relationship between linear and angular motion is highlighted, emphasizing that linear speed (v) is related to angular speed (ω) through the equation v = ωR. The discussion also notes the importance of establishing a coordinate system for analyzing the motion, with the x-axis aligned with the ramp and the z-axis perpendicular to it. Additionally, it clarifies that all vector quantities involved have positive components in this context. The problem is effectively solved using these principles.
Reshma
Messages
749
Reaction score
6
A drum of radius R rolls down a slope without slipping. Its axis has acceleration 'a' parallel to the slope. What is the drum's angular acceleration? Please help me solve in polar coordinates.
 
Physics news on Phys.org
Reshma said:
A drum of radius R rolls down a slope without slipping. Its axis has acceleration 'a' parallel to the slope. What is the drum's angular acceleration? Please help me solve in polar coordinates.
since the drum rolls without slipping:
Angular Acceleration = α = a/R

to understand this, what's the relationship between the linear motion of the drum's axis (moving down the ramp) to the rolling outer surface circumference? remember, there's no slipping.
 
Last edited:
geosonel said:
since the drum rolls without slipping:
Angular Acceleration = α = a/R

to understand this, what's the relationship between the linear motion of the drum's axis (moving down the ramp) to the rolling outer surface circumference? remember, there's no slipping.

Linear speed v = \omega R
So if T is the time period, distance in one revolution = circumference
vT = 2\pi R

Please note that the motion here is downhill(clockwise) so there will be changes in the sign if we take into account the direction. So how do I make these changes?
 
Reshma said:
Linear speed v = \omega R
So if T is the time period, distance in one revolution = circumference
vT = 2\pi R

Please note that the motion here is downhill(clockwise) so there will be changes in the sign if we take into account the direction. So how do I make these changes?
remember that linear velocity (v), linear acceleration (a), angular velocity (ω), and angular acceleration (α) are all vector quantities. when solving a problem, one of the first jobs is to select convenient (orthogonal) coordinate axes into which these vector quantities can be projected into their coordinate components.

for your problem, it's typical to consider the ramp with highest level on the left, lowest level on the right, with motion from "left-to-right". then it's also typical to establish a coordinate system consisting of:
1) "x" axis parallel to ramp surface, (+) direction left-to-right
2) "z" axis normal (perpendicular) to ramp surface, (+) direction upward from ramp surface
3) "y" axis into page (normal to both above axes), (+) direction into page

for your problem, the linear motion is "left-to-right" along the "x" axis and thus its x-component is (+). the angular motion is clockwise, which according to the "right-hand rule", produces a (+) "y" component.
thus, for this problem, all quantities have (+) values for their components.

with the above in mind, and working with the aformentioned components, you basically have solved the problem.
beginning where you left off:

If T is the period of revolution, distance in 1 revolution = 1 circumference:

vT \ = \ 2\pi R

v \ = \ (2\pi /T) R

v \ = \ (\omega) R

\frac{dv}{dt} \ = \ a \ = \ \frac{d(\omega R)}{dt} \ = \ \frac{d\omega}{dt} \cdot R \ = \ \alpha R
 
Last edited:
EDIT: eh crap... made a mistake and your solution right I think
 
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...
Back
Top