Explanation:
(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.
Sum of forces in the centripetal direction:
∑F = ma
mg = mv²/RL
v = √(g RL)
(b) Energy is conserved.
EE = KE + RE + PE
½ kd² = ½ mv² + ½ Iω² + mgh
kd² = mv² + Iω² + 2mgh
kd² = mv² + (m RC²) ω² + 2mg (2 RL)
kd² = mv² + m RC²ω² + 4mg RL
kd² = mv² + mv² + 4mg RL
kd² = 2mv² + 4mg RL
kd² = 2m (v² + 2g RL)
d² = 2m (v² + 2g RL) / k
d = √[2m (v² + 2g RL) / k]
You may know linear momentum is given by
P= mass.velocity.
Initially car is moving with some velocity so you know initial momentum of the car. Finally it comes to rest i.e final momentum of the car is 0. According to Newton's second law : Force = change in momentum /time. Applying this you'll get answer as 642840N. Hope it helped you. Revert back to me if you have any questions. Please check out the calculation it might be wrong!
The answer to this statement is true!
Answer:
The maximum height reached by the ball is 16.35 m.
Explanation:
Given;
initial velocity of the ball, u = 17.9 m/s
the final velocity of the ball at the maximum height, v = 0
The maximum height reached by the ball is given by;
v² = u² + 2gh
During upward motion, gravity is negative
v² = u² + 2(-g)h
v² = u² - 2gh
0 = u² - 2gh
2gh = u²
h = u² / 2g
h = (17.9)² / (2 x 9.8)
h = 16.35 m
Ttherefore, the maximum height reached by the ball is 16.35 m.
