Question

1 S = 1 x 10-13 sectomie R= 8.314 J molt K M =8.314 Nm molt K1 =8.314 kg m? s? mol K-=8.314 x 10'g cm? s? molt k? 1. For a cylinder, the moment of inertia is I = Vmr?, and the kinetic energy of rotation is E = Vlwa. Consider an ultracentrifuge rotor to be a cylinder of titanium (density of 4.5 g/cm, diameter of 25 cm, and height of 10 cm). Calculate the kinetic energy of such a rotor when it is spinning at 50,000 rpm. How fast would you have to drive a one-tonne truck (in km/h) to acquire the equivalent kinetic energy (E = mv2)?

Answer 1

For a cylinder of of radius r and height h, volume is given by

V = arh

Radius is given as r=25/2 cm, height h=10 cm.

25 :: V=a* (?) x 10 cc = 4908.74 ce

Given, density of titanium is 4.5 gm/cc

So, mass of the cylinder is

m = Volume x Density = 4908.74 x 4.5 gm = 22089.32 gm = 22.09 kg

So, moment of inertia of the cylinder is

w By, (1972) X0072 x = wants

It is spinning at 50000 rpm

So, angular speed is

T w = 50000 rpm = 50000 x rad/s

So, rotational kinetic energy is

$E=\frac{1}{2}I\omega^2=\frac{1}{2}\times\frac{1}{2}\times22.09\times\left ( \frac{25}{2\times100} \right )^2\times\left (50000\times\frac{2\pi}{60} \right )^2\ J$

2.37 x 106 1

Let the truck is to be driven at a speed of v m/s

Mass of the truck is

M =1 tonne = 1000 kg

So, to have the same rotational kinetic energy,

Mv2 = 2.37 x 10

or, X 1000 X v2 = 2.37 x 10

or, v= 1 2 x 2.37 x 106 2 m/s = 68.85 m/s

68.85 x 60 x 60 - km/hr = 247.85 km/hr 1000