The figure below (Figure 1) illustrates an Atwood\'s machine. (NEED PARTS A-E) P

Question

The figure below (Figure 1) illustrates an Atwood's machine. (NEED PARTS A-E) PLEASE

***PLEAS NOTE THIS HAS BEEN ANSWERED INCORECTLY ALREADY...

***ANSWERS ARE NOT (A= 2.32, B = 2.32, C=17.24, D= 24.24,E = 26.18)

PART A;

Let the masses of blocks

A and B be 3.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.300 kg?m2, and the radius of the wheel be 0.130 m . Find the magnitude of linear acceleration of block A if there is no slipping between the cord and the surface of the wheel.

a = _____________m/s2

PART B:

Find the magnitude of linear acceleration of block

B if there is no slipping between the cord and the surface of the wheel.  

a =________________m/s2

PART C:

Find the magnitude of angular acceleration of the wheel C if there is no slipping between the cord and the surface of the wheel.

? =______________rad/s2

PART D:

Find the tension in left side of the cord if there is no slipping between the cord and the surface of the wheel.

TB = ____________ N

PART E:

Find the tension in right side of the cord if there is no slipping between the cord and the surface of the wheel.

TA =

_________N

igure below (Figure 1 illustrates an Atwoods ne Submil My Answers Give Up Part C Find the magnitu cord and the surface of the wheel Part D Find the tension in eft side of the cord it there is no slipping

Explanation / Answer

Let’s assume that block A is the left block and block B is the right block.
The weight of block A is greater than the weight of block B. So block A will move downward and block B will move upward. Since the cord is not slipping, the wheel will rotate counterclockwise.
The tension in the left side of the cord is greater than the tension in the right side of the cord, because the weight of block A is greater than the weight of block B.

The tension in the left side of the cord is causing the pulley to rotate counter clockwise. The tension in the right side of the cord is causing the pulley to rotate clockwise.
Torque = Tension * radius, radius = 0.13 m
Counter clockwise torque = T1 * 0.13
Clockwise torque = T2 * 0.13
Net counterclockwise torque = (T1 – T2) * 0.13

Net counterclockwise torque = Moment of inertia * angular acceleration
Net counterclockwise torque = 0.300 *

(T1 – T2) * 0.13 = 0.3 *
(T1 – T2) = 0.44 *

Angular acceleration = Linear acceleration ÷ r
= a ÷ 0.13

(T1 – T2) = 0.44 * a ÷ 0.13
Eq#1: (T1 – T2) = 3.3846 * a


For the blocks the weight is pulling downward and the tension is pulling upward.

For 3.5 kg block, (3.5 * 9.8) – T1 = 3.5 * a
T1 = 34.3 – 3.5 * a


For 2.0 kg block, T2 – (2.0 * 9.8) = 2.0 * a
T2 = 19.6 + 2 * a

T1 – T2 = (34.3 – 3.5 * a) – (19.6 + 2 * a)
Eq#2: T1 – T2 = 14.7 – 5.5 * a


3.3846 * a = 14.7 – 5.5 * a
8.8846 * a = 14.7
a = 1.6545 m/s^2

T1 = 34.3 – 3.5 * 1.6545 = 28.509
T2 = 19.6 + 2 * 1.6545 = 22.909

Angular acceleration = linear acceleration ÷ radius
Angular acceleration = 1.6545/0.13 = 12.7269 rad/s^2

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