Angular Acceleration of Solid Cylinder: Mass, Force, Time

[melissa_y]

1) A solid cylinder of mass M = 1.14 kg and radius R = 6.3 cm pivots on a thin, fixed, frictionless bearing (see Figure). A string wrapped around the cylinder pulls downward with a force F = 4.606 N. What is the magnitude of the angular acceleration of the cylinder? Use units of "rad/s\^{}2".

2)Consider that instead of the force F, a block with mass m = 0.47 kg (with force F = 4.606 N) is attached to the cylinder with a massless string (see Figure). What is now the magnitude of the angular acceleration of the cylinder? Use units of "rad/s\^{}2".

3)How far does mass m travel downward between t = 0.51 s and t = 0.71 s (Assuming motion begins at time t = 0.0 s)?

 

[Dr.Brain]

1)

$T=I a where T = rF sinQ$

Use the expression for MI of the cylinder and calculate the torque and put the values in the above expression.

2) Try to use the Free Body Diagram , the same expressions as above will be applicable , in this case the force 'F' will change.Try i out and show your work here.

B.J

 

[thepatientmental]

1) To find the angular acceleration of the cylinder, we can use the equation α = τ/I where α is the angular acceleration, τ is the torque, and I is the moment of inertia. In this case, the torque is equal to the force F multiplied by the radius R, and the moment of inertia for a solid cylinder is given by I = ½MR². Plugging in the given values, we get:

α = (4.606 N)(0.063 m)/(½)(1.14 kg)(0.063 m²)

= 0.0487 rad/s²

2) If we now attach a block to the cylinder with a massless string, the torque will be the same as before, but the moment of inertia will change. We can calculate the new moment of inertia by adding the moment of inertia of the cylinder to the moment of inertia of the block, which is given by I = ½mr². So, the new moment of inertia is:

I = ½(1.14 kg)(0.063 m²) + ½(0.47 kg)(0.063 m²)

= 0.0387 kgm²

Using the same equation as before, we get:

α = (4.606 N)(0.063 m)/(0.0387 kgm²)

= 0.0750 rad/s²

3) To find the distance traveled by the block between t = 0.51 s and t = 0.71 s, we can use the equation s = ut + ½at² where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time. We can assume that the block starts from rest, so u = 0. The acceleration can be found by multiplying the angular acceleration by the radius, so a = αR. Plugging in the values, we get:

s = 0 + ½(0.0487 rad/s²)(0.063 m)(0.71 s)² - ½(0.0487 rad/s²)(0.063 m)(0.51 s)²

= 0.000797 m

Therefore, the mass m travels 0.000797 m downward between t = 0.51 s and t = 0.71 s.