need help solving and understanding. on the graph it tells us Time=3.36 and Omeg

Question

need help solving and understanding. on the graph it tells us Time=3.36 and Omega=3.293

Physics 2514, Rotational Kinetic Energy The total kinetic energy of a rigid rotating object is the kinetic energy due to the velocity of the center of mass of the object (KEcmS M V2) and the kinete energy due to rotation about the center of mass This discussion explores the rotational kinetic energy A) Consider a wheel of radius R and mass M that is free to rotate about its center All of the mass is on the rim of the wheel If the wheel is rotating with an angular velocity , what is the velocity of the mass in the wheel? What is the kinetic energy of the wheel? B) The rotational kinetic energy of a solid object is written as where "I" depends on the size, shape, mass, and rotation axis for the object. From your result from part (A), what is I for the wheel? C) Consider 4 different wheels all rotating with the same angular velocity, but with different mass and radius: (M, R), (2M, R), (M, 2R), (2M, R/2). Rank in order from largest to smallest the kinetic energy of the wheels. Explain the physics of your answer sturt withdiferent DI A solid disk has an I CMoment of Inertia"l equal to

Explanation / Answer

part A:

as angular velocity=linear velocity/radius

==>linear velocity=v=angular velocity*radius=w*R

kinetic energy=0.5*M*v^2=0.5*M*w^2*R^2

part B:

comparing rotational kinetic energy expression with kinetic energy expression received from part A,

0.5*I*w^2=0.5*M*w^2*R^2

==>I=M*R^2

part C:

as angular velocity is same, larger the value of "I", larger the kinetic energy

I1=M*R^2

I2=2*M*R^2

I3=M*(2*R)^2=4*M*R^2

I4=2*M*(R/2)^2=0.5*M*R^2

hence I3>I2>I1>I4

part D:

when the object is at height h and with speed=0,

total energy=potential energy of the object+potential energy of the pulley

=m*g*h+M*g*H

ii)

when the object is at height h/2, total energy is conserved

hence total energy=m*g*h+M*g*H

in terms of variable at that instant,

if speed is v , then angular speed=v/R

total energy=potential energy of the box+kinetic energy of the box+potential energy of the pulley+rotational kinetic energy of the pulley

==>total energy=m*g*(h/2)+0.5*m*v^2+M*g*H+0.5*I*(v/R)^2

iii) right before hitting ground,height of the object=0
total energy is conserved

hence total energy=m*g*h+M*g*H

in terms of variable at that instant,

if speed is v , then angular speed=v/R

total energy=potential energy of the box+kinetic energy of the box+potential energy of the pulley+rotational kinetic energy of the pulley

==>total energy=0+0.5*m*v^2+M*g*H+0.5*I*(v/R)^2=0.5*m*v^2+M*g*H+0.5*I*(v/R)^2

part E;

given that m=0.5 kg, M=2 kg, R=1.5 m, h=21.5 m,H=26 m

I=0.5*M*R^2=2.25 kg.m^2

then total energy=m*g*h+M*g*H=614.95 J

from part iii of part D, we get when just before hitting ground,

total energy=0.5*m*v^2+M*g*H+0.5*I*(v/R)^2=0.5*0.5*v^2+2*9.8*26+0.5*2.25*(v^2/1.5^2)=509.6+0.25*v^2+0.5*v^2=509.6+0.75*v^2

equating the two expressions,

614.95=509.6+0.75*v^2

==>v=11.8518 m/s

velocity of the object just before hitting ground is 11.8518 m/s

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