ENGINEERING ADMISSIONS ASSESSMENT PREPARATION
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Question 58

4a. $\frac{1}{2}mv^2=mgH$
$v=\sqrt{2gH}$

4b. $\frac{1}{2}mv^2=mg(H-2R)$
$v=\sqrt{2g(H-2R)}$

4c. Circular motion requires a force towards the centre of the circle - think of planetary motion - there is no way the force could be in any other direction.

4d. If the car is to just make it round the loop. not only does it need to have enough gravitational energy to do so, it also needs some velocity to stay on the track. In this case, the reaction force at the top would be zero and the only force acting on the car would be the gravitational force. Equating this to $\frac{mv^2}{r}$ in the question (the force required to sustain circular motion) gives $v=\sqrt{gR}$ but from earlier in the question we have that $v=\sqrt{2g(H−2R)}$. Equating these yields $R=2H-4R$ or $H=2.5R$ and the height needs to be greater than this because a reaction force of exactly zero means that the car has left the track.
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  • Home
  • Tutor profile
  • Solutions
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  • Contact
  • Time saving
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  • Specimen solutions
  • 2016 Solutions
  • 2017 Solutions
  • 2018 Solutions
  • 2019 Solutions
  • 2020 Solutions
  • 2021 Solutions
  • New Specimen Solutions
  • NSAA Specimen Section 2 solutions
  • NSAA 2016 Section 2 Solutions
  • NSAA 2017 Section 2 Solutions
  • NSAA Tuition and eBook
  • PAT Tuition
  • Physics Aptitude Test Mock Papers
  • MAT Tuition
  • STEP Tuition
  • Maths Challenge and BMO Tuition
  • Advance Preparation Programme
  • Physics Degree Tuition
  • Maths enrichment tuition
  • Maths for Economics Degrees Tuition
  • GDPR and data protection
  • Boiler