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allsm [11]
2 years ago
15

5. A hollow cylinder of mass m, radius Rc, and moment of inertia I = mRc2 is pushed against a spring (with spring constant k) co

mpressing it by a distance d. It is then released and rolls without slipping on a track, and through a vertical loop of radius RL. Assume RC << RL
(a) When the cylinder reaches the top of the vertical loop, what is the minimum (linear) speed it must have to avoid falling off? Draw a free-body diagram to support your answer.
(b) What is the minimum compression of the spring necessary to prevent the cylinder falling off?
(c) Perform two “Cross-checks” on you solutions to check the validity of your solution, and/or examine the behavior of the system.

Physics
1 answer:
Makovka662 [10]2 years ago
8 0

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]

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Answer:

m v^2 / R = G M m / R^2       gravitational attraction = centripetal force

M = v^2 R / G       solving for M

period = 6 h 25 min = (6 * 3600 + 25 * 60) sec = 23,100 sec = T

v = 2 pi R / T

M = 4 pi^2 R^3 / (G T^2)

M = 39.5 * (8.6E7)^3 / (6.67E-11 * 2.31E4^2)

M = 39.5 * 636 / (6.67 * 5.34) * 10^24

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8 0
2 years ago
Three 7 0hm resistors are connected in series across a 10 V battery. What is the equivalent resistance of the circuit?
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Given

Three 7 ohm resistor are in series.

The battery is V=10V

To find

The equivalent resistance

Explanation

When the resistance are in series then the resistance are added to find its equivalent.

Thus the equivalent resistance is:

R=7+7+7=21\Omega

Conclusion

The equivalent resistance is 21 ohm

4 0
10 months ago
g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
statuscvo [17]

Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

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If Juan upgrades from a sports car to a large truck, what will happen to the
vovikov84 [41]

Answer:

The force will have to increase

Explanation:

Since Juan has upgraded from a sports car to a large truck, based on Newton's second law of motion, the force needed to keep the truck going at the same speed will have to increase.

 According to Newton's second law "the force on an object is equal to the product of its mass and acceleration".

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A truck has a larger mass compared to a sports car.

By virtue of this, to make sure both automobiles attain the same speed, the force powering them to accelerate must be the same.

 Therefore, the force from the engine must increase.

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