1 1)What is magnitude of the linear acceleration of the hoop? 2)What is magnitud

Question

1

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 = 1.52 m. (After being released from rest.)

How much time does the hoop take to fall 1.52 m?

8)What is the magnitude of the velocity of the green hoop after it has dropped 1.52 m?

9)What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.52 m)?

An green hoop with mass mh-2.5 kg and radius Rh 0.16 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 - 3.4 kg and radius Rs = 0.22 m. The system is released from rest.

Explanation / Answer

Given:-

For Hoop

Mass of hoop = 2.5kg

Radius of hoop = 0.16m

For Solid disk Key,

Mass of disk = 1.9 kg

Radius of disk= 0.1m

Orange Sphere,

mass of orange sphere= 3.4 kg

Radius of orange sphere= 0.22m

"F=ma" equations for each of the things that moves. Also, since some of the objects (the pulley and the orange sphere) rotate, you should write " = I"

equations (net torque = moment of inertia × angular acceleration) for those.

force acting on hoop

F net = 2.5kg*9.81m/s^2 = 24.525 N

equivalent mass of sphere =
So using the theorem of parallel axes,
the moment of inertia of the sphere
is I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I= mR^2(2/5+1)
I = 7/5*mR^2
M = 7/5*m -------------------------------->i {where m= mass of sphere}

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
torque = F*R = I* = I*a/R { where =angular accelaration, I=moment of inertia}
F = M*a = I*a/R^2 -->
M = I/R^2 = 1/2*m*R^2/R^2 = 1/2*m ------> ii { where m-mass of pulley}   

the acceleration is then
a = F/m = 24.525/(2.5 + 1/2*1.9 + 7/5*3.4) = 24.525/3.861 = 6.351 m/s^2
----------------
1)linear acceleration of hoop= a=6.351 m/s^2

2) linear acceleration of sphere = 6.351 m/s^2

3) angular acceleration disk pulley:
= a/R = 6.351/0.1 = 63.51 rad/s^2

4) angular acceleration sphere = a/R = 6.351/0.22 = 28.86 rad/s^2

5) Tension between pulley and sphere = M*a = 7/5*3.4*6.351 = 30.2307 N

6) tension between hoop and pulley = m(hoop) (g - a)
= 2.5(9.81 - 6.351) = 8.64 N

7) s = 1/2*at^2 --> t = (2s/a) = (2*1.52/6.351)
--> t = 0.6918 seconds

8) velocity of greev hoop = (2as) = (2*6.351*1.52) = 4.39 m/s

9) final angular speed of orange sphere() = v/R = 4.39/0.22 = 19.95 rad/s

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