Mass and Kinetic Energy
Disk With Weight:
A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 3 kg weight is hanging on the string. The system of the weight and disk is released from rest.
a) When the 3 kg weight is moving with a speed of 2.2 m/s, what is the kinetic energy of the entire system? KETOT = KEwheel+KEweight = (1/2)(I)(w2)+(1/2)(m*v2) =(0.5* v2)(m+1/2M) =0.5*(2.2^2)*(3+(.5*15)) J
b) If the system started from rest, how far has the weight fallen? H = KETOT/MG = 0.5*(2.2^2)*(3+(.5*15))/(3*9.8) m
c) What is the angular acceleration at this point? Remember that a = αR, or α = a/R
Solve for acceleration by using vf2=vi2+2ax (vf=2.2, vi=0, x=(answer in part b)) That gives you the linear a… we want angular acceleration so we just divide our linear acceleration by the radius:
((2.2)^2)/(2*0.5*(2.2^2)*(3+(.5*15))/(3*9.8))/0.25
Sphere on Incline:
A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 3.7 m down a θ = 33° incline. The sphere has a mass M = 4.2 kg and a radius R = 0.28 m.
For this one, you have to do part © first… and don’t forget that ur working with a sphere, so u gotta look up the moment of inertia (I) in order to get the correct answer…
a) Of the total kinetic energy of the sphere, what fraction is translational? U just divide the translational found in part (b) by KEtotal ((1/2)mv2 + (1/2)Iω2) 0.71
b) What is the translational kinetic energy of the sphere when it reaches the bottom of the incline? (1/0.7)*9.8*(3.7sin(33))*.5*4.2
c) What is the translational speed of the sphere as it reaches the bottom of the ramp? You can find the velocity by using the conservation of energy theorem thing: KEi + Ui = KEf + Uf h = d*sin(θ) Mgh = KEtrans + KErot = (1/2)mv2 + (1/2)Iω2
A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 3 kg weight is hanging on the string. The system of the weight and disk is released from rest.
a) When the 3 kg weight is moving with a speed of 2.2 m/s, what is the kinetic energy of the entire system? KETOT = KEwheel+KEweight = (1/2)(I)(w2)+(1/2)(m*v2) =(0.5* v2)(m+1/2M) =0.5*(2.2^2)*(3+(.5*15)) J
b) If the system started from rest, how far has the weight fallen? H = KETOT/MG = 0.5*(2.2^2)*(3+(.5*15))/(3*9.8) m
c) What is the angular acceleration at this point? Remember that a = αR, or α = a/R
Solve for acceleration by using vf2=vi2+2ax (vf=2.2, vi=0, x=(answer in part b)) That gives you the linear a… we want angular acceleration so we just divide our linear acceleration by the radius:
((2.2)^2)/(2*0.5*(2.2^2)*(3+(.5*15))/(3*9.8))/0.25
Sphere on Incline:
A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 3.7 m down a θ = 33° incline. The sphere has a mass M = 4.2 kg and a radius R = 0.28 m.
For this one, you have to do part © first… and don’t forget that ur working with a sphere, so u gotta look up the moment of inertia (I) in order to get the correct answer…
a) Of the total kinetic energy of the sphere, what fraction is translational? U just divide the translational found in part (b) by KEtotal ((1/2)mv2 + (1/2)Iω2) 0.71
b) What is the translational kinetic energy of the sphere when it reaches the bottom of the incline? (1/0.7)*9.8*(3.7sin(33))*.5*4.2
c) What is the translational speed of the sphere as it reaches the bottom of the ramp? You can find the velocity by using the conservation of energy theorem thing: KEi + Ui = KEf + Uf h = d*sin(θ) Mgh = KEtrans + KErot = (1/2)mv2 + (1/2)Iω2