The rotational energy stroed in a 1.2m radius, uniform-disk flywheel of mass 50k

Question

The rotational energy stroed in a 1.2m radius, uniform-disk flywheel of mass 50kg is being used to lift a spherical object (with a mass of 80kg and radius of 13cm) out of a large vessel of water. WE'll ignore the fluid resistance. The objet is initially submerged by 12m below the water surface, and is needed to be brought above the surface by an additional 6.0m.

A. What is the density of the spherical object?

B. What is the buoyant force acting on the object when it is submerged?

C. How much energy is required to lift the object by the amount specified above?

D. What must be the minimum rotational speed (in revolutions per minute) of the flywheel in order to provide the energy needed in part c?

E. If the cable lifting the mass wraps around a spindle (of radius 0.050m) that supports the flywheel, through how many revolutions does the flywheel turn in lifting the mass the required amount?

Explanation / Answer

ANSWER: A-------> Density of spherical object= mass/volume. mass= 80kg. volume= (4/3)*(pie)*(radius)*(radius)*(radius) pie=22/7 , radius= 0.13m therefor, volume = (4 * 22 * 0.13 * 0.13 * 0.13)/(3 * 7 ) density= mass/volume =(80)/((4 * 22 * 0.13 * 0.13 * 0.13)/(3 * 7 )) answer=8693.03 kg/meter.cube ANSWER B-: Buoyant force acting on shpere = volume of sphere*density of water volume of sphre = (4 * 22 * 0.13 * 0.13 * 0.13)/(3 * 7 ) density of water = 1000 kg/meter.cube bouyant force = (4 * 22 * 0.13 * 0.13 * 0.13)/(3 * 7 ) * 1000 N =9.202 Newton ANSWER C -: energy required to lift bring object till 12m inside water =E1 = (m*g - bouyant force) * height. =(80*9.8 - 9.202) * 12 Joule =9297.58 Joule work done till 6m = E2 = mgh =80* 9.8*6 = 4704 joule total energy = E1 + E2 14001.58 joule ANSWER D -: Energy= 1/2 I w2 where, I = the moment of inertia of the object about an axis through the centre of gravity of the system w = the angular velocity of the system about an axis through the centre of gravity of the system. E= 1/2 I w2 for disc I=1/2 M R2 I= 1/2 * 50* (1.2*1.2) =36 kg/m2 E= 1/2 I w2 14001.58 joule = 1/2 *36 * w2 w= 27.89 must me the angular speed of disc. ANSWER E -: Total distance required to cover = 18m distance in one rotation of wheel = 2* pie * R =2* 22/7 * 0.05 =0.314 18=n* 0.314 N=57.33 times or we can say 54 revolution is required.

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