Use height and speed without importing a circular-motion force model.
Syllabus 4.5
Notes and key methods
For a particle constrained to a smooth fixed guide in a vertical plane, the normal constraint force does no work. If gravity is the only other force, kinetic plus gravitational potential energy is constant.
A vertical drop h increases speed squared by 2gh, regardless of the intermediate path. To find a later highest accessible point, use zero speed and account for the original kinetic energy as well as its height. Only overall energy changes are needed here; no centripetal-force equation is required.
Check the conditions
Choose and retain a positive direction; distinguish signed quantities from magnitudes.
Use SI units and \(\displaystyle g= \frac{10m}{s^{2}}\) wherever gravity is relevant.
Apply only the model and time interval stated in the question.
Three worked examples
Example 1 · Energy on a smooth curved track
A bead of mass 9 kg is constrained to a smooth fixed guide in a vertical plane. At A its speed is \(\displaystyle \frac{5m}{s}\). It descends a vertical height \(\displaystyle \frac{119}{20}\) m to B, then the guide rises far enough for it to reach its first stop. Gravity is the only force doing work, and the bead remains on the guide. Use \(\displaystyle g= \frac{10m}{s^{2}}\).
Find its speed at B in m/s.
Find its kinetic-energy increase from A to B in J.
Find how far above A it can rise before its first stop, in m.
Find the greatest vertical height above B in m.
Show the reasoning
(a)
Convert the gravitational energy loss to kinetic-energy gain and use the positive root.
Conservation gives \(\displaystyle v^{2}=u^{2}+20h=144\), so the speed is \(\displaystyle \frac{12m}{s}\).
Answer: \(12\).
(b)
Use conservation of mechanical energy between the two specified heights.
The lost gravitational potential energy is Mgh=\(\displaystyle \frac{1071}{2}\) J.
Answer: \(\displaystyle \frac{1071}{2}\).
(c)
Set final speed to zero while retaining the initial kinetic energy at A.
Its initial kinetic energy supports extra rise \(\displaystyle \frac{u^{2}}{2g}=\)\(\displaystyle \frac{5}{4}\) m above A.
Answer: \(\displaystyle \frac{5}{4}\).
(d)
Use B as the energy datum for the final rise to zero speed.
All kinetic energy at B converts to potential energy over height \(\displaystyle \frac{v^{2}}{2g}=\)\(\displaystyle \frac{36}{5}\) m.
Answer: \(\displaystyle \frac{36}{5}\).
Example 2 · Energy on a smooth curved track
A bead of mass 3 kg is constrained to a smooth fixed guide in a vertical plane. At A its speed is \(\displaystyle \frac{6m}{s}\). It descends a vertical height \(\displaystyle \frac{133}{20}\) m to B, then the guide rises far enough for it to reach its first stop. Gravity is the only force doing work, and the bead remains on the guide. Use \(\displaystyle g= \frac{10m}{s^{2}}\).
Find its speed at B in m/s.
Find its kinetic-energy increase from A to B in J.
Find how far above A it can rise before its first stop, in m.
Find the greatest vertical height above B in m.
Show the reasoning
(a)
Convert the gravitational energy loss to kinetic-energy gain and use the positive root.
Conservation gives \(\displaystyle v^{2}=u^{2}+20h=169\), so the speed is \(\displaystyle \frac{13m}{s}\).
Answer: \(13\).
(b)
Use conservation of mechanical energy between the two specified heights.
The lost gravitational potential energy is Mgh=\(\displaystyle \frac{399}{2}\) J.
Answer: \(\displaystyle \frac{399}{2}\).
(c)
Set final speed to zero while retaining the initial kinetic energy at A.
Its initial kinetic energy supports extra rise \(\displaystyle \frac{u^{2}}{2g}=\)\(\displaystyle \frac{9}{5}\) m above A.
Answer: \(\displaystyle \frac{9}{5}\).
(d)
Use B as the energy datum for the final rise to zero speed.
All kinetic energy at B converts to potential energy over height \(\displaystyle \frac{v^{2}}{2g}=\)\(\displaystyle \frac{169}{20}\) m.
Answer: \(\displaystyle \frac{169}{20}\).
Example 3 · Energy on a smooth curved track
A bead of mass 4 kg is constrained to a smooth fixed guide in a vertical plane. At A its speed is \(\displaystyle \frac{7m}{s}\). It descends a vertical height \(\displaystyle \frac{147}{20}\) m to B, then the guide rises far enough for it to reach its first stop. Gravity is the only force doing work, and the bead remains on the guide. Use \(\displaystyle g= \frac{10m}{s^{2}}\).
Find its speed at B in m/s.
Find its kinetic-energy increase from A to B in J.
Find how far above A it can rise before its first stop, in m.
Find the greatest vertical height above B in m.
Show the reasoning
(a)
Convert the gravitational energy loss to kinetic-energy gain and use the positive root.
Conservation gives \(\displaystyle v^{2}=u^{2}+20h=196\), so the speed is \(\displaystyle \frac{14m}{s}\).
Answer: \(14\).
(b)
Use conservation of mechanical energy between the two specified heights.
The lost gravitational potential energy is Mgh=\(294\) J.
Answer: \(294\).
(c)
Set final speed to zero while retaining the initial kinetic energy at A.
Its initial kinetic energy supports extra rise \(\displaystyle \frac{u^{2}}{2g}=\)\(\displaystyle \frac{49}{20}\) m above A.
Answer: \(\displaystyle \frac{49}{20}\).
(d)
Use B as the energy datum for the final rise to zero speed.
All kinetic energy at B converts to potential energy over height \(\displaystyle \frac{v^{2}}{2g}=\)\(\displaystyle \frac{49}{5}\) m.
Answer: \(\displaystyle \frac{49}{5}\).
Common mistakes
Use the vertical height drop, not an unspecified curved path length.
Initial kinetic energy allows the particle to rise above its original height.