Answer: 4.98 m/s
Explanation:
You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.
PEi = 0 in this case
KEi = ½mVi² = PEf+KEf = mghf + ½mVf²
½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²
½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26
Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s
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:
The correct answer is
a) 1, 2, 3
Explanation:
In rolling down an inclined plane, the potential energy is Transferred to both linear and rotational kinetic energy thus
PE = KE or mgh = 1/2×m×v² + 1/2×I×ω²
The transformation equation fom potential to kinetic energy is =
m×g×h = 
= 
= 
=
Therefore the order is with increasing rotational kinetic energy hence
the first is the sphere 1 followed by the disc 2 then the hoop 3
the correct order is a, 1, 2, 3
The answer is D using the work formula
W= F•d but if it was against gravity, it would be 0 if gravity is exerting the same amount, I would pick D using the formula, but I'm not so sure sorry
