Question

(1) The earth's radius is 6.37×106m; it rotates once every 24 hours.

(A) What is the speed of a point on the equator?

(2) The (Figure 1) shows two thin beams joined at right angles. The vertical beam is 13.0 kg and 1.00 m long and the horizontal beam is 28.0 kg and 2.00 m long. Neglect the lateral dimensions of the beams.

(A) Calculate the gravitational torque on the joined beams about an axis through the corner.

(3) A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin shell with much higher density. A 7.0 lb (3.2 kg) bowling ball has a diameter of 0.216 m; 0.196 m of this is a 1.6 kg core, surrounded by a 1.6 kgshell. This composition gives the ball a higher moment of inertia than it would have if it were made of a uniform material. Given the importance of the angular motion of the ball as it moves down the alley, this has real consequences for the game.

(A) Compare the moments of inertia for these models by finding the ratio of Ir and Iu.

Answer 1

1) radius of the earth R= 6.37*10^6 m

   angular speed = 2N    where N is the number of rotations per second

   given N = 1 rotation in 24 hours

  = 2N = 2(1/24*60*60) = 2(1/86400)

speed of a point at the equator V = R

   V =(6.37*10^6)2(1/86400)

   V = 463.24 m/s

2)

  

3)Given mass of uniform solid spherr would be 3.2 kg

   mass of lower density sphere =1.6 kg

   mass of higher density hollow sphere =1.6 kg

moment of inertia of uniform solid sphere I_u= 2/5 MR^2 =(2/5) (3.2)(0.216)^2=0.0597 kg m^2

moment of inertia of irregular sphere = moment of inertia of inner core + moment of inertia of outer core

   moment of inertia of inner solid spher =2/5 (m)R^2 = (2/5)(1.6)(0.196)^2 =0.0246 kgm^2

   moment of inertia of outer hollow sphere = 2/5m(0.216^-0.196^2) = (2/3)(1.6)(0.00824)=0.0053kgm^2

   momemt of inertia of irregular sphere I_r =0.0246+0.0053=0.0299 kgm^2

I_u/I_r = 0.0597/0.0299 = 1.996 = 2

   I_u =2I_r