Circular Motion on a Hemisphere: Finding Speed from Geometric Conditions

AI Thread Summary
The discussion revolves around calculating the speed of an object in uniform circular motion on a hemisphere with a radius of r, positioned at a height of r/5 from the lowest point. The user initially derived equations using forces and geometry but encountered discrepancies in the final speed calculation. After analyzing the geometry of the hemisphere, the correct tangent value was identified as 3/4, leading to the correct speed formula. The final speed was confirmed to be (3√5gr)/10 after correcting the geometric conditions. The thread highlights the importance of accurate geometric interpretation in solving physics problems.
IIK*JII
Messages
21
Reaction score
0

Homework Statement


As shown in attached figure, a small object is in uniform circular motion in a horizontal plane, on the smooth of a hemisphere (radius:r). The distance between the object's plane of motion and the hemisphere's lowest point is \frac{r}{5}

What is the speed of the object?


Homework Equations


ƩF=\frac{mv<sup>2</sup>}{r} ...(1)

Ncosθ=mg ...(2)



The Attempt at a Solution



From (1)
and I get ƩF from FBD wrote in attached figure ƩF=Nsinθ

From (2) I knew that N=\frac{mg}{cosθ} ..(3)

substitute (3) in (1) in got gtanθ=\frac{v<sup>2</sup>}{r} ..(4)

and I try to find tanθ from geometric of hemisphere

First, I try to find the radius (let it is r') of this mass at r/5 from the lowest point of hemisphere

If I look in the picture and use pythagoras r' = (r2-(\frac{4r}{5}))1/2
∴r' = \frac{3r}{5}
Thus; tanθ = 3
substitute in (4) v = √3gr

but the answer is \frac{3√5gr}{10}...

Or I get tanθ wrong or use wrong geometric condition of hemisphere ?

help is appreciate

Thanks :!)
 

Attachments

  • hemisphere2.JPG
    hemisphere2.JPG
    19.2 KB · Views: 726
Physics news on Phys.org
Your tanθ should be equal to (3/5)r/(4/5)r=3/4

tanθ=v2/r'g
v2=3/4x3/5rg=3/4x3/5grx5/5=3.3.5gr/4.5.5
 
Oh! thanks azizlwl :wink:

i get it this is an easy one but i can't notice o:)
 
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