Solve Angular Motion Problems: Work, Inertia & Force

[Destrio]

1. A rigid body, starting at rest, rotates about a fixed axis with a constant angular acceleration α. Consider a particle a distance r from the axis. Express (a) the radial acceleration and (b) the tangential acceleration of this particle in terms of α, r and time t.
c) if the resultant acceleration of the particle at some instant makes an angle of 57.0 degrees with the tangential acceleration, through what total angle has the body rotated from t=0 to that instant.

i got radial acceleration = rω^2 = (α^2)(t^2)r
tangential acceleration = αr

for c i made a triangle and got cos(57deg) = cos(.99rad) = (α^2)(t^2)r/αr = αt^2

so i plugged it into the kinematics formula
theta = (1/2)αt^2
theta = (1/2)cos(.99rad)
theta = .27 radians
but that is incorrect

2. Two particles, each with mass m, are fastened to each other and to a rotation axis by two rods, each with length L and mass M. The combination rotates around the rotation axis with angular velocity ω. Obtain an algebraic expression for the rotational inertia of the combination about the axis.

I = m1r1^2 + m2r2^2
I = ML^2 + M(2L)^2
I = 5ML^2
im not sure any other way to solve this

3. A 52.3-kg trunk is pushed 5.95m at constant speed up a 28.0 degree incline by a constant horizontal force. The coefficient of kinetic friction between the trunk and the incline is .19 . Calculate the work done by a) the applied force and b) the force of gravity.

I found the forge of gravity one easily, just mgh

for a) I figured I should make my axes line up with the slope so
Sum of forces in x = Fcos28 - f - mgsin28 = 0
and then solving for F

thanks

 

[anathema]

for the help

For problem 1, your expressions for radial acceleration and tangential acceleration are correct. For part c, you made a small mistake in your calculation. The correct expression for the angle would be theta = (1/2)cos^-1(αt^2), and plugging in the values would give you an angle of approximately 0.27 radians, or 15.5 degrees.

For problem 2, your expression for the rotational inertia is correct. Another way to solve it would be to use the parallel axis theorem, which states that the rotational inertia of a system is equal to the rotational inertia of an object about its center of mass plus the product of its mass and the square of the distance between the center of mass and the axis of rotation.

For problem 3, your approach is correct. The work done by the applied force can be calculated using W = Fd, where F is the applied force and d is the displacement along the incline. The work done by the force of gravity can be calculated using the equation you mentioned, mgh. Remember to use the component of the force of gravity parallel to the incline, which is mg sin 28 degrees.

 

[ogb p]

for your question!

To address the first problem, your calculations for the radial and tangential accelerations are correct. However, for part c, you need to use the formula for angular displacement, not kinematics. The correct equation to use is θ = ωt + (1/2)αt^2. Plugging in the values, we get:

θ = (αt^2)/2 = (.99/2)t^2

To find the angle of rotation, we can plug in the given values for time (t) and angular acceleration (α) and solve for θ:

θ = (.99/2)(5)^2 = 12.375 radians or approximately 0.21 revolutions.

For the second problem, your calculation for the rotational inertia is correct. Another way to solve it would be to use the parallel axis theorem, which states that the moment of inertia of a body about any axis is equal to the moment of inertia about a parallel axis through the center of mass plus the product of the mass and the square of the distance between the two axes. In this case, the parallel axis would be through the center of mass of the combination of particles. So the equation would be:

I = Icm + Md^2

Where Icm is the moment of inertia about the center of mass, and d is the distance between the two axes (in this case, the length of the rods). Plugging in the values, we get:

I = (mL^2 + m(2L)^2) + M(2L)^2
I = 5mL^2

For the third problem, you are correct in setting up the equation for the sum of forces in the x-direction. To find the work done by the applied force, we can use the equation W = Fd, where F is the applied force and d is the displacement in the direction of the force. In this case, the displacement is along the incline, so we need to use the component of the force (Fcos28) in that direction. So the work done by the applied force would be:

W = Fcos28 * 5.95m = 40.04 Joules

To find the work done by the force of gravity, we need to use the equation W = mgh. In this case, the height (h) would be the vertical displacement of the trunk, which is h = 5