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]
Answer:
0.1 N
Explanation:
Considering the relationship between force,
spring constant and extension as defined by Hook's law
The force F=xk as from Hooke's law where F is the force of the spring, k is spring constant and x is extension or compression. Substituting 2 N/m for k and 5cm which is equivalent to 0.05 m for extention x then the force will be
F=2*0.05=0.1 N
Answer:
The correct one is that the force on B is half of the force on A
Explanation:
Because radius for the inside of the curve is half the radius for the outside and Car A travels on the inside while car B, travels at equal speed on the outside of the curve. Thus force on B will be half on A
