Angular acceleration of a cylinder rolling up an inclined plane?

[khfrekek92]

Homework Statement

The cylinder is being pulled up the plane by a block. I've found that it IS rolling without slipping, and the acceleration is 5.00156 m/s^2.

Homework Equations

a=R[alpha]

The Attempt at a Solution

I've heard that the angular acceleration is the linear acceleration/radius, but this doesn't make sense, because using a radius of .2m, I get 25.0078 s^-2. What kind of units are inverse seconds squared? Does an angular acceleration of 25 even make sense for this situation? and is that even the right units? I thought it was supposed to be in rad/sec^2?

I've also seen that [torque]=I[alpha] Can I use that somehow? What would the torque be?

 

[khfrekek92]

Now I've tried using [torque]=rf and [torque]=I[alpha] and set them equal to each other and have gotten [alpha]=42.44.. How can this be right?? I'm getting two completely different answers :( :(

 

[Dick]

You haven't really described the whole problem so I don't know why you are getting two answers for the torque. But a radian angular measure is the distance along a circle divided by the radius of the circle. Since they are both measured in meters the units of radians are meters/meters. There's nothing wrong with writing rad/sec^2 but since 'rad' is formally dimensionless, they often omit writing 'rad'.

 

[khfrekek92]

Okay awesome! Well there are two inclined planes connected at the tips each with angle [theta] and [phi] which are 30 degrees, and 60 degrees respectively. On the 30 degree plane is a cylinder with mass 1.2 kg and radius .2m, and on the 60 degree plane is a block of mass 3 kg. These two objects are connected by a string running over a pulley. So the block is pulling the cylinder up its plane. The tension in the string is 16.97 N, and the acceleration (linear) is 5.00156 m/s^2. I'm not used to angular acceleration so I'm not sure if 20 rad/s^2 is acceptable for this situation... :/

 

[rwisz]

I would first congratulate the individual on their findings and their understanding of the concept of rolling without slipping. However, I would also suggest that they revisit their calculations and units to ensure accuracy.

To address their concerns about the units of inverse seconds squared, I would explain that angular acceleration is typically measured in radians per second squared (rad/s^2), which is a unit of rotational acceleration. It is the rate at which the angular velocity of an object changes over time.

In this situation, the angular acceleration can be calculated using the equation a=R[alpha], where a is the linear acceleration, R is the radius of the cylinder, and [alpha] is the angular acceleration. So, using a radius of 0.2m and a linear acceleration of 5.00156 m/s^2, we can find the angular acceleration to be 25 rad/s^2.

To address their question about using torque, I would explain that torque is the rotational equivalent of force, and it can be calculated using the equation [torque]=I[alpha], where [torque] is the torque, I is the moment of inertia, and [alpha] is the angular acceleration. The moment of inertia depends on the shape and mass distribution of the object, so it would need to be determined in order to use this equation.

In conclusion, the angular acceleration of 25 rad/s^2 seems reasonable for this situation, and it can be calculated using either the equation a=R[alpha] or [torque]=I[alpha]. It is important to use the correct units and to double-check calculations to ensure accuracy.