Centripetal acceleration of viking ship in the amusement park.

In summary: Hello, There is no diagram in the original question. Otherwise, I would have reproduced it here. Thanks for your response.
  • #1
WMDhamnekar
MHB
376
28
Hi,

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity.

(a) Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system’s center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass.

(b) What is the centripetal acceleration at the bottom of the arc?

(c) Draw a free body diagram of the forces acting on a rider at the bottom of the arc.

(d) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.

(e) Discuss whether the answer seems reasonable.

Answers:- $(a)16.57 m/s (b) 19.61 m/s^2$

(c)
1604473600989.png
N=North pole= Upward forces acting on the rider at the bottom of arc. W=?I know it is downward forces $Mass \times g(9.81 m/s^2)$

But what are your answers to (d) and (e)?
 
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  • #2
Is there a diagram with this problem? What is the position of the ride relative to its lowest position when it is momentarily motionless?
 
  • #3
skeeter said:
Is there a diagram with this problem? What is the position of the ride relative to its lowest position when it is momentarily motionless?
Hi,
I got the answer for (d) and it is 1176.798 N or $2.00 \times W$ where $W= 60 kg \times 9.80665 m/s^2$
 
  • #4
Dhamnekar Winod said:
Hi,
I got the answer for (d) and it is 1176.798 N or $2.00 \times W$ where $W= 60 kg \times 9.80665 m/s^2$

Is that your solution, or one given by "the book"?

by the way, you didn't answer my first question ...

Is there a diagram with this problem? What is the position of the ride relative to its lowest position when it is momentarily motionless?
 
  • #5
Hi,
Question states 60 kg rider but give answer for 90kg rider. :)
 
  • #6
You still have not answered my first question ...

If the normal force on the rider is $2mg$, then the ship makes a 60 degree angle w/respect to the vertical at its highest position.
 
  • #7
skeeter said:
You still have not answered my first question ...

If the normal force on the rider is $2mg$, then the ship makes a 60 degree angle w/respect to the vertical at its highest position.
Hello,
How did you compute that angle?
 
  • #8
Dhamnekar Winod said:
Hello,
How did you compute that angle?

I used right triangle trigonometry.
 
  • #9
skeeter said:
I used right triangle trigonometry.
Hello,
That means force exerted by the ride on the rider can't be more than $3mg_r$
 
  • #10
Dhamnekar Winod said:
Hello,
That means force exerted by the ride on the rider can't be more than $3mg_r$

That's right ... as I stated in post #6, if the angle the ship makes with the vertical at its highest position is 60 degrees, then the normal force exerted on the rider is $2mg$.

You still haven't posted the original problem or any related diagram as I requested earlier in this thread. Please do so or this will be my final response.
 
  • #11
skeeter said:
That's right ... as I stated in post #6, if the angle the ship makes with the vertical at its highest position is 60 degrees, then the normal force exerted on the rider is $2mg$.

You still haven't posted the original problem or any related diagram as I requested earlier in this thread. Please do so or this will be my final response.
Hello,
There is no diagram in the original question. Otherwise, I would have reproduced it here. Thanks for your response.
 

Related to Centripetal acceleration of viking ship in the amusement park.

What is centripetal acceleration?

Centripetal acceleration is the acceleration towards the center of a circular path. It is caused by a net force acting towards the center of the circle.

How is centripetal acceleration calculated?

Centripetal acceleration can be calculated using the formula a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path.

Why does the viking ship in the amusement park experience centripetal acceleration?

The viking ship in the amusement park experiences centripetal acceleration because it is constantly changing direction as it moves along its circular path. This change in direction requires a centripetal force to keep the ship from moving in a straight line.

What factors affect the centripetal acceleration of the viking ship?

The centripetal acceleration of the viking ship is affected by the speed of the ship, the radius of the circular path, and the mass of the ship. A higher speed or smaller radius will result in a greater centripetal acceleration, while a heavier ship will require a greater centripetal force to maintain its circular motion.

How does centripetal acceleration impact the riders on the viking ship?

Centripetal acceleration can cause riders on the viking ship to feel a force pushing them towards the center of the circle. This can result in a sensation of weightlessness or a feeling of being pressed against the seat, depending on the direction of the acceleration. If the acceleration is too strong, it can also cause discomfort or even nausea in some riders.

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