Pop-up Toy Physics Situation

Question

Please help.

An inventor has created a spring-loaded toy that pops up off the table when compressed and released. The toy consists of two connected parts by a stiff spring as shown, with one of the parts being much heavier than the other. For which orientation, A or B, will the toy rise farther off the table after being compressed and released?

Thanks.

Answer 1

The full analysis of this type of problem is fairly complicated and I don't have access to the figure you mention.

However, I can provide some insight. The first point is that there is a minimum threshold value, C_min , of compression that is required to get the bottom part to budge at all. C_min has dimensions of length.

Below, I will show that C_min = (m / M) (M + m) g / k if the small mass part is on the top and

C_min = (M/ m) (M + m) g/k if the large mass part is on the top. Here M and m are the masses of the heavy and light parts respectively , g = 9.81 and k is the spring constant. From this it is clear that C_m is much lower if the small mass part is on the top. Naturally, the larger is the applied compression relative to C_min , the higher the toy will rise above the table. Further, it will be mechanically easier to apply a large C with the light mass on top without the two masses coming into contact. Taken together, it is seen the the small mass on top configuration will be the better design.

The final part of this response is a derivation of the result above: C_min = (m/M) ( M + m) g/ k.

for the light mass on top design.

Consider a compression C, small enough such that the mass on the bottom does not move. Under this condition, the upper mass will execute simple harmonic motion (SHM) with frequency ω = sqrt( k/ m). For this SHM, the coordinate of the top part is given by x - x₀ = - C cos (ω t) , where x₀ is the equilibrium point about which x oscillates. It can be worked out that the coordinate y, of the center of mass is given by

y - y₀ = - C (M/ (M + m) ) cos(ω t). The acceleration of the center of mass is the second derivative of this expression : a_CM = C ω² ( M/(M + m)) cos (ω t) = C (k/m) (M / (M+m)) cos(ωt ).

By Newton's second law F_CM = (M +m) a_CM = C k (M/m) cos(ω t)

We are most interested at the top-of-the-motion point where cos (ω t) = -1, This is the point where the downward force on the CM is its largest. The net force on the CM is F_CM = N - (M + m) g, where N is the normal force exerted by the table on the system

Putting this together (for cos(ω t) = -1 and solving for N gives

N = (M + m ) g - Ck (M/m) . Since N must be greater than or equal to zero for the bottom mass not to move, this gives the condition C_min = (m/M) (M + m) g/k.

The same analytic approach gives C_min = (M/m) (M + m) g/k for the heavy mass on top design.