Question
An green hoop with mass mh = 2.4 kg and radius Rh = 0.14 m hangs from a string that goes over a blue solid disk pulley with mass md = 2.2 kg and radius Rd = 0.1 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass ms = 3.5 kg and radius Rs = 0.22 m. The system is released from rest.
1)What is magnitude of the linear acceleration of the hoop?
2)What is magnitude of the linear acceleration of the sphere?
3)What is the magnitude of the angular acceleration of the disk pulley?
4)What is the magnitude of the angular acceleration of the sphere?
5)What is the tension in the string between the sphere and disk pulley?
6)What is the tension in the string between the hoop and disk pulley?
7)The green hoop falls a distance d = 1.7 m. (After being released from rest.)How much time does the hoop take to fall 1.7 m?
8)What is the magnitude of the velocity of the green hoop after it has dropped 1.7 m?
9)What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.7 m)?
Someone please help me! I'm so lost!
1)What is magnitude of the linear acceleration of the hoop?
2)What is magnitude of the linear acceleration of the sphere?
3)What is the magnitude of the angular acceleration of the disk pulley?
4)What is the magnitude of the angular acceleration of the sphere?
5)What is the tension in the string between the sphere and disk pulley?
6)What is the tension in the string between the hoop and disk pulley?
7)The green hoop falls a distance d = 1.7 m. (After being released from rest.)How much time does the hoop take to fall 1.7 m?
8)What is the magnitude of the velocity of the green hoop after it has dropped 1.7 m?
9)What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.7 m)?
Someone please help me! I'm so lost!
Answers
PIAFMSPI
HOW DO YOU DO THIS>>>
Carly
1) F=ma
a=(mh*g)/(7/5ms+1/2md+mh)
2) same answer as 1 because it is a system
3)a=alpha*R
alpha=a/Rd (use a from 1)
4) similar to 3
alpha=a/Rs
5)F=ma
T=ms*a
T=7/5ms*a (a from 1)
6)T=mh(g-a)
7)d=1/2at^2 (position formula - starting position and velocity are zero)
t=sqrt(2d/a) (use a from 1)
8)vf^2=vi^2 +2ad (initial velocity = 0)
vf=sqrt(2ad)
9)omega=v/Rs
a=(mh*g)/(7/5ms+1/2md+mh)
2) same answer as 1 because it is a system
3)a=alpha*R
alpha=a/Rd (use a from 1)
4) similar to 3
alpha=a/Rs
5)F=ma
T=ms*a
T=7/5ms*a (a from 1)
6)T=mh(g-a)
7)d=1/2at^2 (position formula - starting position and velocity are zero)
t=sqrt(2d/a) (use a from 1)
8)vf^2=vi^2 +2ad (initial velocity = 0)
vf=sqrt(2ad)
9)omega=v/Rs
Anonymous
no you
anonomos
bro im literally going to end it this is not a joke this is a serious cry for help
anon
Why am I going into a physics field why did I do this to myself
bruh
physics makes me bruh
crying
were tf did the guy get 7/5 from, that's the one thing I just have no understanding whatsoever of
Popsmokeontop
Moment of intertia
Moises Valle
why is the only force acting on the system is the hoop but the tension is not included?
Elon Musk
Y'all got this! Keep up the good work, I need more engineers :)