How Do You Calculate Speed and Normal Forces in a Loop-the-Loop Problem?

Click For Summary
SUMMARY

The discussion focuses on calculating speed and normal forces for a block sliding through a frictionless loop-the-loop. The block, with a mass of 3.0 kg, is released from a height of 2.9R (where R = 0.80 m) and the calculations involve converting potential energy (PE) at the top to kinetic energy (KE) at the bottom. The normal forces at various points, including the bottom and top of the loop, require applying the total force equation F_total = ma along with centripetal acceleration principles. Key equations utilized include KE = 1/2mv² and the conservation of energy principle.

PREREQUISITES
  • Understanding of kinetic energy (KE) and potential energy (PE) concepts
  • Familiarity with centripetal acceleration and its application in circular motion
  • Knowledge of Newton's second law (F = ma)
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Study the conservation of mechanical energy in physics problems
  • Learn about centripetal force and its role in circular motion
  • Explore detailed examples of normal force calculations in different scenarios
  • Investigate the effects of friction in loop-the-loop problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to enhance their understanding of energy conservation and forces in circular motion.

gcharles_42
Messages
11
Reaction score
0

Homework Statement



In Fig. 8-28, a small block of mass m = 3.0 kg can slide along the frictionless loop-the-loop. The loop has radius R = 0.80 m. The block is released from rest at point P, at height h = 2.9R above the bottom of the loop.

I attatched a picture

(a) Find the speed of the block when it reaches point Q.

(b) Find the normal force on the block at point Q.

(c) Find the normal force on the block when it is at the very bottom.

(d) Find the normal force on the block when it is at the top of the little loop (i.e., at a height 2R above the ground)




Homework Equations



KEi + PEi = KEf +PEf

KE=1/2mv^2

v = w (r)

The Attempt at a Solution



So for the velocity isn't it just converting the PE at the top for the KE at the bottom?

For all the normal forces i am completely stumped
 

Attachments

  • Screen Shot 2013-10-18 at 4.44.08 PM.png
    Screen Shot 2013-10-18 at 4.44.08 PM.png
    2.6 KB · Views: 663
Physics news on Phys.org
So for a) v = 3(.98) 1.9(.8) = 1/2(3)v^2
 
hi gcharles_42! :smile:
gcharles_42 said:
So for the velocity isn't it just converting the PE at the top for the KE at the bottom?

yes

but i don't understand the LHS of your equation :redface:
gcharles_42 said:
So for a) v = 3(.98) 1.9(.8) = 1/2(3)v^2

for the normal forces, you need to use Ftotal = ma in the normal direction

(you'll need the centripetal acceleration formula)
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 12 ·
Replies
12
Views
12K
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
31
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K