Help Solving Rotational Motion Homework Questions

[ghadz8]

Homework Statement

Can someone help me those 2 questions

Homework Equations

I don't understand how to convert to angular acceleration

The Attempt at a Solution

I attempted of finding the torque due to gravity then divided it by the radius for the second question, I got it wrong, the correct answer was supposed to be 27.8 N.

for the first question I did something similar I got 1.4 which is the correct answer but I'm not sure if my methods was correct because all I did was try and balance the dimensions.

PS . I kept the rules of the forum, I got my homework wrong and the teacher gave the rigth answers, I just don't know how exactly to start these kind of problems.

 

[ehild]

You need the keep the rules of the Forum. Show your attempt to solve the problem first.

ehild

 

[HallsofIvy]

But you still have not show any work. Can you find the center of mass of the first object? What is its distance from the pivot?

 

[tiny-tim]

welcome to pf!

hi ghadz8! welcome to pf!

for the first question I did something similar I got 1.4 which is the correct answer but I'm not sure if my methods was correct because all I did was try and balance the dimensions.

show us exactly what you did, so that we can comment

 

[ghadz8]

Question 1:
T = mg(a+0.5c)costheta = I alpha ( a is the length in the picture )
mg(a+0.5c0costheta = 5 x 9.81 x cos(30) = 27.61 N m
I = (1/12)m(b^2+c^2) + mr^2= (1/12) x 5 x (0.2^2 + 0.3^2) + 5 (0.65)^2 = 2.167 kg m^2
Alpha = 12.744 rad/s^2
Atangential = alpha x r = 12.744 x 0.65 = 8.283 m/s^2
Aradial = w^2 x r = 3^2 x 0.65 = 5.85 m/s^2
I’m stuck here, how do I find the force? I’m supposed to get 27.8 N according to the answers but I’m not quite sure how to get there from this step.

Question 2:
For this question I assumed:
Atangential =gcostheta = 9.81 x cos (60) = 4.5 m/s^2
Alpha = a/r = 4.5 / 3.5 = 1.4 rad/s^2
My answer was correct but could someone confirm that my working out was right and it wasn’t a coincidental answer.

 

[ghadz8]

Could you people at least just give me a little hint? thanks

 

[ehild]

Question 1:
T = mg(a+0.5)costheta = I alpha

Do you mean T = mg(0.5a +c) cos (theta)?

The centre of mass of the pendulum moves along a circle with the given angular speed, under the effect of gravity and the radial force from the peg. What should be the radial component of the resultant force? The CM also has tangential acceleration corresponding to the resultant of the tangential force components. The tangential acceleration equals the angular acceleration times the radius of the circle, from that you get the force component which is normal to the rod.

ehild

 

[ghadz8]

What do you call a?

The centre of mass of the pendulum moves along a circle with the given angular speed, under the effect of gravity and the force from the peg. What should be the radial component of the resultant force?

ehild

a is given in the picture, I meant a + 0.5 x c

 

[ehild]

Question 2:
For this question I assumed:
Atangential =gcostheta = 9.81 x cos (60) = 4.5 m/s^2
Alpha = a/r = 4.5 / 3.5 = 1.4 rad/s^2
My answer was correct but could someone confirm that my working out was right and it wasn’t a coincidental answer.

This is all right.

ehild

 

[ghadz8]

This is all right.

ehild

Can someone direct me for question 2, that's the hard one and tell me which step is wrong, at least help me arage the formula for forces

 

[tiny-tim]

hi ghadz8!

(have an alpha: α and a theta: θ and a degree: ° and try using the X² and X₂ icons just above the Reply box )

Question 2:
For this question I assumed:
Atangential =gcostheta = 9.81 x cos (60) = 4.5 m/s^2
Alpha = a/r = 4.5 / 3.5 = 1.4 rad/s^2
My answer was correct but could someone confirm that my working out was right and it wasn’t a coincidental answer.

with questions like this, always start with good ol' Newton's second law F = ma …

first you must decide in which direction to apply it (it's valid in every direction, but it's always quicker if you choose the direction perpendicular to any unknown force, since the component of that force will then be zero) …

in this case, what is the unknown force, and so in which direction should you do F = ma ? ​

 

[ehild]

You completely neglected the centripetal acceleration.

ehild

 

[tiny-tim]

You completely neglected the centripetal acceleration.

ehild

that's the idea!

 

[ehild]

Question 1:
Atangential = alpha x r = 12.744 x 0.65 = 8.283 m/s^2
Aradial = w^2 x r = 3^2 x 0.65 = 5.85 m/s^2

Sorry, I did not notice the radial acceleration.. Your values are correct.

So you have both components of the acceleration of the CM. Collect the forces and apply Newton's second law for both the tangential and radial directions, as Tiny-Tim suggested. You have G=mg and the force of the pin (N). Both have radial and "tangential" components. So you have the equations:

m a_radial=G_radial + N _radial,
m a_tangential= G_tangential + N_tangential.

At the end, calculate the vertical components of the force of the pin and add them. You will get the the correct solution.

ehild