Answer to Question #281892 in Physics for Fahad

Question

A pinball machine launch ramp consisting of a spring and a 30∘ ramp of length L.

(a) (3 marks) If the spring is compressed a distance 𝑥 from its equilibrium position and is then released at 𝑡 = 0, the pinball (a sphere of mass 𝑚 and radius 𝑟) reaches the top of the ramp at 𝑡 = 𝑇. Derive the expression for the spring constant 𝑘 in terms of 𝑚, 𝑔, 𝑥, and 𝐿.[Assume that the friction is sufficient, and the ball begins rolling without slipping immediately after launch.]

(b) (2 marks) What is the potential energy of the ball when it is at the midpoint of the ramp?

(c) (3 marks) Derive the expression of the speed of the ball immediately after being launched in terms of 𝑔 and 𝐿. 

Answer 1

(a) Derive the expression for the spring constant using energy conservation principle:

$$\frac12kx^2=mgL\sin\theta,\\\space\\ k=\frac{2mgL\sin\theta}{x^2}=\frac{mgL}{x^2}.$$

(b) The potential energy at the midpoint:

$$E_P=mg\frac H2=\frac12mgL\sin\theta=\frac14mgL.$$

(c) The expression of the speed of the ball immediately after being launched:

$$\frac12kx^2=\frac12mv^2+\frac12I\omega^2,\\\space\\ \frac12kx^2=\frac12mv^2+\frac12\bigg(\frac25mr^2\bigg)\bigg(\frac vr\bigg)^2,\\\space\\ \frac12kx^2=\frac12mv^2+\frac15mv^2,\\\space\\ v=x\sqrt{\frac {5k}{7m}}.$$