Monday
March 27, 2017

Post a New Question

Posted by on .

A 3.00 kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform 8.00 kg cylindrical pulley. The inner diameter of the pulley is 60.0 cm, while the outer diameter is 1.00 m. The system is released from rest, and there is no friction at the axle of the pulley. Find (a) the acceleration of the stone, (b) the tension in the wire, and (c) the angular acceleration of the pulley.

for part a i did
I of the pulley is
I = m*(r1^2 + r2^2)/2)
I = 8*(0.5^2 + 0.3^2)/2)
I =1.36 kgm^2

a = 3*9.81/(3 + 1.36/0.5^2)
a = 3.48 m/s^2 <--- acceleration of the stone

part B
รก = a/R = 3.48/0.5 = 6.96 rad/s^2 <--- angular acceleration

Part C
T = mg - ma = 3(9.81-3.48)=18.99 N <--- tension

  • Physics please help! check if its correct - ,

    B and C is switched anwsers

  • Oh my - sign ! - ,

    The inner radius I assume means that the axle does not spin. Therefore your moment of inertia has a negative, not positive sign
    I = 8*(0.5^2 - 0.3^2)/2)
    I =.64 kgm^2

  • Physics please help! check if its correct - ,

    stone F = m a
    T up
    mg down
    a = (mg-T)/ m = (3*9.81 -T)/3

    wheel Torque = I alpha
    alpha = .5 T/.64
    a = alpha * r
    a = .25 T/.64

    so
    (3*9.81 -T)/3 = .25 T/.64

    solve for T
    go back and solve for a
    and for alpha = a/r

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question