an atwoods machine has a solid cylindrical pulley with a mass of 200g and a radius of 10cm. a 1.2 kg mass is hung on one side and a 1.1kg mass on the other side, find the acceleration of the two masses and the itme for them to travel a distance of 60cm
Model the pulley as a solid disk, and get the moment of inertia for that model.
(1.2kg-1.1kg)g= 1.2kg*a+1.1kg*a+ I*a/r
solve for a
To find the acceleration of the two masses in the Atwood's machine, we can use the following formula:
a = (m2 - m1) * g / (m1 + m2)
where:
a is the acceleration
m1 is the mass on one side (1.2 kg)
m2 is the mass on the other side (1.1 kg)
g is the acceleration due to gravity (9.8 m/s^2)
Substituting the given values into the formula:
a = (1.1 kg - 1.2 kg) * 9.8 m/s^2 / (1.2 kg + 1.1 kg)
This simplifies to:
a = -0.1 kg * 9.8 m/s^2 / 2.3 kg
a = -0.098 m/s^2
The negative sign indicates that the masses will move in opposite directions, with the smaller mass going up and the larger mass going down.
Now, let's calculate the time it takes for them to travel a distance of 60 cm (0.6 m). We can use the following equation:
d = (1/2) * a * t^2
where:
d is the distance traveled (0.6 m)
a is the acceleration (-0.098 m/s^2)
t is the time in seconds (what we're looking for)
Rearranging the equation to solve for t:
t = √(2 * d / a)
Substituting the given values into the equation:
t = √(2 * 0.6 m / -0.098 m/s^2)
This gives us:
t ≈ √(-12.24 s^2/m)
Since the square root of a negative number is imaginary, the time it takes for them to travel a distance of 60 cm is not a real number. This means that the masses will not reach the specified distance in this scenario.