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pishuonlain [190]
3 years ago
14

An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that a person inside is stuck

to the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8 m/s 2 . Given g = 9.8 m/s 2 , the coefficient µ = 0.521 of static friction between a person and the wall, and the radius of the cylinder R = 4.4 m. For simplicity, neglect the person’s depth and assume he or she is just a physical point on the wall. The person’s speed is v = 2πR T where T is the rotation period of the cylinder (the time to complete a full circle). Find the maximum rotation period T of the cylinder which would prevent a 40 kg person from falling down. Answer in units of s.
Physics
1 answer:
ElenaW [278]3 years ago
7 0

Let N be the normal force that forces the person against the wall.

Then u N = m g  is the frictional force supporting the person's weight

and N = m g / u

also, N = m v^2 / R is the normal force providing the centripetal acceleration

So, m g / u = m v^2 / R

v^2 = g R / u

since v = 2 pi R T

4 pi^2 R^2 T^2 = g R / u     and T^2 = g / (4  u pi^2 R)

T = 1/ (2 pi)  (g /(u R))^1/2 = .159 * (9.8 m/s^2 / (.521 * 4.4 m)) ^1/2

T = .68 / s

Do you see any thing wrong here?

T should have units of seconds not 1 / seconds

v should be  2 * pi * R / T  where T is the time for 1 revolution

So you need to make that correction in the above formula for v.

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Dahasolnce [82]
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Define moment of momentum. at which condition is it's magnitude zero?​
ololo11 [35]

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However, photons are considered to have a rest mass zero.

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Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

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8 0
3 years ago
A 1000 kg satellite and a 2000 kg satellite follow exactly the same orbit around the earth. What is the ratio F1/F2 of the gravi
Tamiku [17]

Answer:

the <em>ratio F1/F2 = 1/2</em>

the <em>ratio a1/a2 = 1</em>

Explanation:

The force that both satellites experience is:

F1 = G M_e m1 / r²       and

F2 = G M_e m2 / r²

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • r is the orbital radius
  • M_e is the mass of Earth

Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

F1/F2 = 1000/2000

<em>F1/F2 = 1/2</em>

The other force that the two satellites experience is the centripetal force. Therefore,

F1c = m1 v² / r    and

F2c = m2 v² / r

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
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  • r is the orbital velocity

Thus,

a1 = v² / r ⇒ v² = r a1    and

a2 = v² / r ⇒ v² = r a2

Therefore,

F1c = m1 a1 r / r = m1 a1

F2c = m2 a2 r / r = m2 a2

In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,

F1 = F1c

G M_e m1 / r² = m1 a1

a1 = G M_e / r²

also

a2 = G M_e / r²

Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

<em>a1/a2 = 1</em>

4 0
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