Respuesta :
Answer:
 3.66 m /s
Explanation:
Initially the ball is at a height of 1.151m ,so its potential energy
= mx9.8x1.151 = 11.2798m J
If V be its required velocity there then its kinetic energy = 1/2 m V²
Total mechanical energy = 11.2798m + 1/2 m V²
The potential energy at the top of loop =
= m x 9.8 x 2 x .7350 =14.406m
if v be the velocity there , the centripetal force must balance the weight there so mv²/r = mg ; v² = gr
kinetic energy = 1/2 m v² = 1/2 m gr = .5 x m x 9.8 x .7350 = 3.6m
Applying conservation of energy ,
11.2798m + 1/2 m V²  = 14.406m +3.6 m
11.2798 + .5 V² = 14.406 + 3.6
V² = 13.4524
V = 3.66 m/s
Answer:
v = 2.0201 m/s
Explanation:
Given:
- mass of the solid m = 0.715 kg
- Radius of the track r = 0.735 m
- Height H Â = 1.151 m
Find:
- What minimum translation speed must the ball have when it is a height H=1.151 m above the bottom of the loop in order to complete the loop without falling off the track?
Solution:
- Draw a free body diagram of the sphere with two forces Normal contact force F_n and Weight of the solid W. We can observe that as the object moves in a circular path its F_n. And W changes with the angle Q it makes with the center point of circular loop. We will apply Newton's Second law of motion in the radial direction:
                      F_n - W = m*a
- At the top of the track the F_n normal contact force vanishes F_n = 0. Hence we have:
                        W = m*a
- The acceleration in this case is the centripetal acceleration a_c which is related to the translation speed as:
                       a_c = v^2 / r
Where, r is the radius of the circular track. Hence we have,
                        m*g*cos(Q) = m*v^2 / r
- Simplify: Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â v^2 = r*g*cos(Q)
                        v = sqrt(r*g*cos(Q))
- Now compute angle Q:
                        cos(Q) = 0.416 / r
- substitute: Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â v = sqrt(0.416*g)
- Plug values in: Â Â Â Â Â Â Â Â Â Â Â v = sqrt(0.416*9.81)
                         v = 2.0201 m/s
- This is the minimum speed required by the object at height H = 1.151 m to complete the entire loop.
