AP Physics C Mechanics - 2.10 Circular Motion- Exam Style questions- FRQs
Question
i. Derive an expression for the speed \(v\) of the block at point \(B\).
ii. Derive an expression for the magnitude of the net force \(F\) on the block at point \(B\).
ii. Explain the reason for the shape of section II on the graph.
Most-appropriate topic codes (AP Physics C: Mechanics):
• Topic 3.4 — Conservation of Energy (Parts b, c, d, e)
▶️ Answer/Explanation
(a)
At point B, the block is changing direction, which requires a centripetal component of acceleration pointing horizontally toward the center of the loop. At the same time, there is a gravitational acceleration acting downward. Because both of these components are present, the net acceleration vector of the block will be directed down and to the left.
(b)(i)
We can find the speed at point B by using the conservation of energy between the release point A and point B.
\(K_{A}+U_{gA}=K_{B}+U_{gB}\)
Setting the initial kinetic energy to zero and substituting the standard expressions gives:
\(\frac{1}{2}mv_{A}^{2}+mgh_{A}=\frac{1}{2}mv_{B}^{2}+mgh_{B}\)
\(0+mgh=\frac{1}{2}mv_{B}^{2}+mgR\)
Solving for \(v_B\) yields:
\(v_{B}=\sqrt{2g(h-R)}\)
(b)(ii)
First, determine the centripetal force using the velocity found in the previous part.
\(F_{c}=\frac{mv_{B}^{2}}{R}=\frac{m\left(\sqrt{2g(h-R)}\right)^{2}}{R}=\frac{2mg(h-R)}{R}\)
The net force is the vector sum of the centripetal force (acting horizontally) and the gravitational force (acting vertically).
\(F_{net}=\sqrt{F_{c}^{2}+(mg)^{2}}\)
Substituting our expression for \(F_c\) yields:
\(F_{net}=\sqrt{\left(\frac{2mg}{R}(h-R)\right)^{2}+(mg)^{2}}\)
(c)
We apply conservation of energy to find the speed of the block at the top of the loop (point C).
\(K_{A}+U_{gA}=K_{C}+U_{gC}\)
\(0+mgh=\frac{1}{2}mv_{C}^{2}+mg(2R)\)
\(v_{C}=\sqrt{2g(h-2R)}\)
For the block to just maintain contact at point C, the normal force approaches zero, meaning gravity alone provides the centripetal force.
\(F_{c}=\frac{mv_{C}^{2}}{R} \implies mg=\frac{mv_{C}^{2}}{R} \implies v_{C}=\sqrt{gR}\)
Equating our two expressions for \(v_C\) gives:
\(\sqrt{gR}=\sqrt{2g(h-2R)}\)
\(R=2(h-2R) \implies \frac{R}{2}=h-2R\)
\(h_{min}=2.5R\)
(d)
We can equate the initial gravitational potential energy at release to the elastic potential energy when the spring is fully compressed.
\(mgh=\frac{1}{2}kx_{MAX}^{2}\)
Substituting the given spring constant \(k=\frac{mg}{2R}\):
\(mgh=\frac{1}{2}\left(\frac{mg}{2R}\right)x_{MAX}^{2}\)
\(x_{MAX}=\sqrt{4hR}\)
Substitute the given numerical values for \(h\) and \(R\):
\(x_{MAX}=\sqrt{4(0.30\,\text{m})(0.10\,\text{m})} = \sqrt{0.12}\,\text{m} \approx 0.35\,\text{m}\)
(e)(i)
For heights less than \(h_{min}\), the block doesn’t have enough speed to make it through the loop. Because it falls off the track, it never reaches the horizontal surface to interact with the spring, resulting in a maximum compression of zero.
(e)(ii)
As established in our derivation for part (d), the maximum compression \(x_{MAX}\) is equal to \(\sqrt{4hR}\). Because the compression is directly proportional to the square root of the initial height \(h\), the graph of compression versus height takes the shape of a square root function (a sideways parabola).