Conservation of Energy with Rotation

AI Thread Summary
The problem involves a circular object rolling without slipping, with a moment of inertia defined by I = cmr^2, launching off a ramp to a maximum height. The initial height is H = 6.0 m, and the ramp height is R = 2.5 m. The user initially calculated the maximum height after launch as 4.7 m, but the textbook states it should be 5.0 m. The confusion arose from an incorrect velocity calculation at the end of the ramp, which was later acknowledged as a mistake due to misreferencing the constant c. Ultimately, the correct approach involves solving symbolically before plugging in numbers, confirming the textbook's answer.
NATURE.M
Messages
298
Reaction score
0

Homework Statement



A small circular object with mass m and radius r has a moment of inertia given by I = cmr^2. The
object rolls without slipping along the track shown in the figure. The track ends with a ramp of height R = 2.5 m that launches the object vertically. The object starts from a height H = 6.0 m. To what maximum height will it rise after leaving the ramp if c = 0.40?


The Attempt at a Solution



My solution to this problem is 4.7 m.
So I'm pretty sure my answer is right but the textbook indicates the answer is 5.0 m which confuses me.

I applied conservation of energy from the starting point, mgH to the launch point (when it leave the ramp), mgR + 1/mv^2 + 1/2I\omega^2 , and then solved for v obtaining 6.55m/s.
Then, I used the equation v^2 = 2g(h-R), where h is the maximum height reached (note I'm
pretty sure once the object leaves the ramp, it loses its rotational motion and maintains only
linear motion after-though correct me if I'm wrong). Rearranging and substituting I obtain
h=4.7m.
 
Physics news on Phys.org
The textbook is correct here. The magnitude of velocity at the end of ramp is not 6.55 m/s.

That said, I suggest that you do not plug in the numbers until you solve everything symbolically. You will see that many things cancel each other out.
 
voko said:
The textbook is correct here. The magnitude of velocity at the end of ramp is not 6.55 m/s.

That said, I suggest that you do not plug in the numbers until you solve everything symbolically. You will see that many things cancel each other out.

Rather silly mistake on my part. I rearranged my equations and everything, but for some reason wrote c= 0.60 at the top of my page and kept referencing that value ...
 
Which, I assume, means you got the correct result?
 
voko said:
Which, I assume, means you got the correct result?

oh yeah I did. Thanks.
 

Thread 'Errors and significant figures'

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...
View full post »
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
View full post »
Thread 'A cylinder connected to a hanging mass'

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...
View full post »

Similar threads

A ball climbing a ramp while "rolling the wrong way"
Replies
32
Views
3K
Loop problem with skier on ramp going into a loop-the-loop
Replies
13
Views
2K
Conservation of Energy when lifting a box up off the floor
Replies
5
Views
2K
What mass must be placed on the cord to keep the pulley from rotating?
Replies
1
Views
873
Two masses on a rotating platform
Replies
3
Views
2K
Energy and momentum conservation
Replies
6
Views
2K
Distance rolled of sphere vs cylinder with same m, r, and v
Replies
5
Views
709
Conservation of energy problem: Ball rolling down inclined plane and then through a loop-the-loop
Replies
10
Views
2K
Sphere rolling down incline then up incline?
Replies
5
Views
2K
Can Energy Conservation Solve the Angular Velocity Problem?
Replies
6
Views
724

Hot Threads

Recent Insights

Back
Top