An green hoop with mass m h = 2.6 kg and radius R h = 0.14 m hangs from a string

Question

An green hoop with mass mh = 2.6 kg and radius Rh = 0.14 m hangs from a string that goes over a blue solid disk pulley with mass md = 1.9 kg and radius Rd = 0.08 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 = 4.2 kg and radius Rs= 0.22 m. The system is released from rest.

1)

What is magnitude of the linear acceleration of the hoop?  

m/s2

2)

What is magnitude of the linear acceleration of the sphere?  

m/s2

3)

What is the magnitude of the angular acceleration of the disk pulley?

rad/s2

4)

What is the magnitude of the angular acceleration of the sphere?  

rad/s2

5)

What is the tension in the string between the sphere and disk pulley?  

N

6)

What is the tension in the string between the hoop and disk pulley?  

N

7)

The green hoop falls a distance d = 1.61 m. (After being released from rest.)

How much time does the hoop take to fall 1.61 m?

s

8)

What is the magnitude of the velocity of the green hoop after it has dropped 1.61 m?  

m/s

9)

What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.61 m)?

rad/s

Explanation / Answer

The only force on the system is the weight of the hoop
Fnet=mhg= 26*9.8= 25.48 N
The inert masses that must be accelerated are the ring, pulley, and the rolling sphere. The mass equivalent of M the pulley is found by
T = FR = I*alpha= I*a/R
F = M*a = I*a/R2
M = I/R2 = (1/2)*m*R2/R2 = 1/2*m
The mass equivalent of the rolling sphere is found from parallel axis theorem
I = 2/5*mR2+ mR2
I = 7/5*mR2
M = 7/5*m
1.

linear acceleration of hoop

a = F/m = 25.48/(2.6 + (1/2)*1.9 + (7/5)*4.2)

a=2.7 m/s2

2.

linear acceleration of the sphere

a=2.7 m/s2

3.

angular acceleration disk pulley:

alpha =a/Rd =2.7/0.08 =33.75 rad/s2

4.

angular acceleration sphere is

alpha =a/Rs =2.7/0.22 =12.27 rad/s2

5)

the tension in the string between the sphere and disk pulley

T =Ma =(7/5)*4.2*2.7 =15.88 N

6.

the tension in the string between the hoop and disk pulley

T=Mh(g-a) =2.6(9.8-2.7) =18.46 N

7.Time taken by hoop to fall is

t=sqrt[2d/a]=sqrt[2*1.61/2.7] =1.09 s

8)

the velocity of the green hoop after it has dropped 1.61 m is

V=sqrt[2ad] =sqrt[2*2.7*1.61] = 2.95 m/s

9)

final angular speed of the orange sphere

W=V/R =2.95/0.22 =13.4 rad/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.