Uniform Circular Motion and Centripetal acceleration

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SUMMARY

The discussion centers on calculating the minimum centripetal acceleration required for a 1,200 kg pickup truck to successfully jump across a 10.0 m wide gully while navigating a circular curve with a radius of 0.333 km. The relevant equations include centripetal acceleration (a(c) = v^2 / r) and the forces involved (Fc = mv^2 / r). The final calculated centripetal acceleration needed for the truck to clear the gully is determined to be 1.08 m/s², emphasizing the importance of accounting for the truck's length in the calculations.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula (a(c) = v^2 / r)
  • Familiarity with Newton's second law of motion (Fc = mv^2 / r)
  • Basic knowledge of kinetic energy (KE = 1/2mv^2) and potential energy (PE = mgh)
  • Ability to apply kinematic equations in projectile motion scenarios
NEXT STEPS
  • Study the principles of projectile motion to understand the gully jump dynamics
  • Learn about the effects of mass and radius on centripetal force in circular motion
  • Explore real-world applications of centripetal acceleration in automotive engineering
  • Investigate the impact of vehicle dimensions on trajectory calculations in stunt driving
USEFUL FOR

Physics students, automotive engineers, stunt coordinators, and anyone interested in the dynamics of motion in circular paths and projectile scenarios.

suyashr99
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Homework Statement


For a scene in a movie, a stunt driver drives a 1.20✕ 103 kg pickup truck with a length of 4.65 m around a circular curve with a radius of curvature of 0.333 km (Fig. 7.29). The truck is to curve off the road, jump across a gully 10.0 m wide, and land on the other side 2.96 m below the initial side. What is the minimum centripetal acceleration the truck must have in going around the circular curve to clear the gully and land on the other side?
upload_2015-11-10_20-28-6.png


Homework Equations


a(c) = v^2/ r
Fc = mv^2/ r
(I don't really know what else to apply here)
I guess we could use energy or kinematics for the ravine part.
KE= 1/2mv^2
PE= mgh

The Attempt at a Solution


0.333 km= 333 m
Fc = mv^2/ r
This is where i got lost and confused.
Please guide me.
 
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Attempt 1: Used the maximum height that it can travel to find the time it would take to cross the gulch (t = 0.77 secs). Divided that by distance to get a velocity (12.9). Plugged that into the formula for centripetal acceleration and got .49 which is wrong.
 
I think you have to take into account the length of the truck, otherwise you are calculating the speed it would take for the truck's front bumper to clear the gully, leaving the rest of the truck to fall in!
 
Mister T said:
I think you have to take into account the length of the truck, otherwise you are calculating the speed it would take for the truck's front bumper to clear the gully, leaving the rest of the truck to fall in!
Okay so i woulld just add right? thanks ill try that.
 
Mister T said:
I think you have to take into account the length of the truck, otherwise you are calculating the speed it would take for the truck's front bumper to clear the gully, leaving the rest of the truck to fall in!
Okay thank you that was correct
 
Hi, i am really interested in this question and would like to attempt this question on my own. So is it possible if you can post the answer to this question? Thank you
 
SWJ said:
Hi, i am really interested in this question and would like to attempt this question on my own. So is it possible if you can post the answer to this question? Thank you
No, you should post your own attempt first.
 
No, i meant the final answer to this question, not the solution so that i know if what i get in the end is correct.
 
The answer is 1.08 m/s^2
 

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