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 = 0.1 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 = 4.1 kg and radius Rs= 0.21 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 = 1.57 m. (After being released from rest.)

How much time does the hoop take to fall 1.57 m? s

8)What is the magnitude of the velocity of the green hoop after it has dropped 1.57 m? m/s

9)What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.57 m)? rad/s

Explanation / Answer

string is connected at centrer and it can not provide angular acc.

so there must be friction force present there on blue sphere.

on green hoop: (Fnet = ma)

2.6g - T1 = 2.6a .......(i)

on blue soild disk,
(troque = I x alpha)

r(T1 - T2) = (m r^2 /2)(a/r)

T1 - T2 = 1.9a/2 = 0.95 .......(ii)

on orange sphere,

T2 - f = 4.1 a ............(iii)

and torque equation,

r f = (2 m r^2 / 5)(a/r)

f = 1.64a ......(iv)

adding all the equations,

2.6g = (2.6 + 0.95 + 4.1 + 1.64)a

a = 2.74 m/s^2

1) a = 2.74 m/s^2

2) a = 2.74 m/s^2

3) alpha = a/r = 2.74 / 0.1 = 27.4 rad/s^2

4) alpha = a/r = 2.74 / 0.21 = 13.06 rad/s^2

5) adding iii and iv,

T2 = (4.1 + 1.64)a = 15.73 N

6) T1 = 2.6 (g - a) = 2.6 x (9.8 - 2.74) = 18.36 N

7) using d = ut + at^2 /2

1.57 = 0 + 2.74t^2 / 2

t = 1.07 s

8) v = u + at

v = 0 + (1.07 x 2.74) =2.93 m/s

9) w = v/r = 2.93 / 0.21 =13.97 rad/s

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