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]
The total energy equation would be Kinetic energy+Potential energy
Answer:
Final angular velocity is 35rpm
Explanation:
Angular velocity is given by the equation:
I1w1i + I2w2i = I1w1f -I2w2f
But the two disks are identical, so Ii =I2
wf can be calculated using
wf = w1i - w2i/2
Given: w1i =50rpm w2i= 30rpm
wf= (50 + 20) / 2
wf= 70/2 = 35rpm
Answer:
40 m/s
Explanation:
given,
height of the fall, h = 82 m
time taken to fall, t = 1.3 s
rock velocity, v = ?
acceleration due to gravity, g = 9.8 m/s²
rock is released initial velocity, u = 0 m/s
using equation of motion
v² = u² + 2 a s
v² = 0 + 2 x 9.8 x 82
v² = 1607.2
v = 40 m/s
hence, rock's velocity is equal to 40 m/s
Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
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