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 $v=\sqrt{2g(Hâˆ’2R)}$. 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.

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 $v=\sqrt{2g(Hâˆ’2R)}$. 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.