- #1

PiEpsilon

- 22

- 2

- Homework Statement
- A pair of ##10cm## radius, ##2kg## disc flywheels with frictionless bearings are spinning at ##1000 rads\space s^-1## and are supported at a distance ##d = 15cm## on either side of a universal bearing by a small diameter bar ##AB## of mass ##M = 1kg## and length ##4d##, as shown in Fig.16-21.

If a ball of mass ##m = 10g## is dropped from a height ##h = 5cm## onto the tip ##A## of the bar and rebounds upwards, give the components of resulting angular momentum ##\vec L## of the flywheels and sketch the motion of the tip of the bar as seen from ##+x## direction. Also, give the angular velocity ##\Omega_n## of this motion and radius ##r## of the circle described by the tip once the circular motion is attained.

- Relevant Equations
- -

What we know:

The ball is dropped at the tip A with some speed ##v_0## and rebounds with speed ##v##. This collision produces an angular impulse, changing the angular momentum of the bar with the flywheels.

*Solution inspired by an answer provided by @TSny in the similar question.*

Angular impulse is:

##m(v + v_0)\cdot 2d = L_y##

By conservation of energy:

##\frac 12m v_0^2 + \frac12 I_x\omega_0^2 = \frac12 mv^2 + \frac12 I_x\omega_0^2 + \frac12 I_y\omega_y^2 \implies m(v_0^2-v^2) = L_y^2/I_y##

Therefore:

##4m^2d^2(v+v_0)^2 = I_ym(v_0^2-v^2)\implies 4md^2(v+v_0) = I_y(v_o-v)##

Solving for ##v## yields:

$$v = \biggr(\frac{I_y-4md^2}{I_y+4md^2}\biggl)v_0\approx(1-8md^2/I_y)\cdot v_0$$

(approximation ignores higher order terms)

##I_y = 2\cdot (\frac14 M_{0}R_{0}^2 + M_{0}d^2) + \frac1{12}M(4d)^2 = (\frac43 M + 2M_{0})d^2 + \frac12 M_{0}R_{0}^2 = 0.13 [kg\cdot m^2]##

*where ##M_0## and ##R_0## is the mass and radius of the disc respectively.*

By the work energy theorem:

##mgh = \frac12 mv_0^2\implies v_0=\sqrt{2gh}\approx 0.99\frac{m}{s}##

Therefore:

##L_y = 2d\cdot mv_0(\frac{I_y-4md^2}{I_y+4md^2}+1) = 2* 0.15 *0.01*\sqrt{2*9.81*0.05}\biggl(\frac{0.13 - 4*0.01*0.15^2}{0.13 + 4*0.01*0.15^2}+1)\biggr)\approx 0.0059 Js^{-1}##

##L_x = I_x\omega_0 = 2\cdot (\frac12M_0R_0^2)\omega_0 = 20Js^{-1}##

And

##\vec L = (20\hat e_x + 0.0059\hat e_y)Js^{-1}##

Problems with the exercise:

The answer of ##L_y## differs by a factor of 2 (should be twice as big).

I assume that ##\Omega_n## is the angular velocity (pointing in

*+ z-direction*) as the bar rotates and aligns with angular momentum vector ##\vec L##.

The angle swept is: ##\lvert\Delta\vec L\rvert/\lvert\vec L\rvert=\lvert L_y\rvert/\lvert\vec L\rvert=\sin\theta##

However in what time does it happen?

I've been trying to use the nutation equation ##\vec\tau = \vec\Omega \times \vec L## too, but I was unable to come up with anything clever.

I am also unable to visualize the tip describing a circular motion. The tip should dip a little bit to conserve angular momentum in the

*z-direction*and rotate about that axis to align itself with the final angular momentum vector ##\vec L##

*.*Considering nutation quickly lead to the mess, that seemed like an overkill for this type of a problem.

Last edited by a moderator: