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skad [1K]
3 years ago
14

Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the

top of the ride?
Physics
1 answer:
masya89 [10]3 years ago
6 0

The complete question is;

In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?

Answer:

F_top = 385.36 N

Explanation:

We are given;

mass;m = 52 kg

Time;t = 4.3 s

Diameter;d = 16m

So,Radius;r = 16/2 = 8m

The formula for the centrifugal force is given as;

F_c = mω²R

Where;

R = radius

Angular velocity;ω = 2πf

f = frequency = 1/t = 1/4.3 Hz

F_c = 53 × (2π × 1/4.3)² × 8 = 905.29 N.

The force at top would be;

F_top = F_c - mg

F_top = 905.29 - (9.81 × 53) N

F_top = 385.36 N

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Two identical blocks are attached to the same massless rope, which is strung around two massless, frictionless pulleys. A massle
Alik [6]

Answer:

The value of mass 1, m1= 6/5m

The value of mass 2, m2= 3/5m

Explanation:

case 1:

here tension and the acceleration will be:

for m1;

  • mg-T=ma
  • 2mg - 2T = 2ma  .....1.

for m2:

  • 2T-mg = ma/2 ..... 2.

adding the both equations,

2mg - 2T + 2T-mg = 2ma + ma/2

a = 2/5 g

putting the value of a into the equation 1.

mg - T = m* (2/5)g

T = 3/5 ( mg )

now

case 2:

The two identical blocks are released from the rest, the tension remains the same as the case 1.

so,

for m1:

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for m2:

  • 2m2g - 2T =0

adding both equations we get,

2T-m2g + 2m2g - 2T = 0

m2 = m1 / 2

T = m1*g / 2

here we know that

T (case1) = T (case2)

3/5 ( mg ) = m1*g / 2

m1 = 6/5 m

hence

m2 = 3/5 m

learn more about tension here:

<u>brainly.com/question/23590078</u>

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#SPJ4

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