Derive and calculate a) the tangential acceleration of an Atwood machine consist

Question

Derive and calculate a) the tangential acceleration of an Atwood machine consisting of two masses m1=1.90kg and m2=2.20 kg suspended from a string passing over a solid cylindrical pulley of mass 5.00 kg and radius 12.0 cm. (I=1/2 MR2 ) b)Find the tensions in the strings. If the smaller mass rests originally on the ground, and the larger mass is located at 4.00m above the ground, c) find the speed of the larger mass just before it hits the ground. d) How high does the smaller mass rise? Indicate forces and torques.

Answer: a) a=0.445m/s2 b) T1=19.5N; T2=20.6N c) v=1.89 m/s d) 4.18m

Explanation / Answer

If m is the mass of the pulley of radius r, its moment of inertia is I. The masses m1 and m2 are pulled by the tensions T1 and T2 respectively.

The second Newton's law for rotation for the pulley is:
net = = I    (1)
That is :

T1 r - T2 r = r(T1 - T2) = I ................. (1')

is the angular acceleration of the pulley.

The second Newton's law for translation of the two masses is:

For m1:
m1 g - T1 = m1 a .................... (4)

For m2:
- m2 g + T2 = m2 a ..................... (5)

From (4):
T1 = m1 g - m1 a = m1 (g - a) ....................... (4')

From (5):
T2 = m2 g + m2 a = m2(g + a) .................... (5')

1.3. Solving for the tensions and accelerations

The tangential acceleration "a1" for the mass m1 is
a1 = r ................ (2)
And
The tangential acceleration "a2" for the mass m2 is
a2 = r ...................... (3)

Therefore
a1 = a2 =a

We have then:
a = r = g (m1 - m2)/((m1 + m2) + I/r2)

T1 = m1(g - a)

T2 = m2(g + a)

a) Putting all values we get a= 0.445 m/s2

b) T1=17.77 N , T2 =22.54 N

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

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