Max Angle for Block to Roll Without Slipping: 29.24°

[Anya91]

1. Homework Statement
A block of a certain material begins to slide on an inclined plane when the plane is inclined to an angle of 15.64°. If a solid cyclinder is fashioned from the same material, what will be the maximum angle at which it will roll without slipping on the plan

2. Homework Equations

tan θ = μ
tan 15.64 = 0.28 = μ
Now let's assume a solid cylinder with mass M and radius R on an inclination θ
Translatory motion, (a is acceleration of CM)
Mg sinθ - μMgcosθ = Ma
g sinθ - μgcosθ = a ...(1)

Rotational motion,
μMgcosθ(R) = Iα (α is angular acceleration)
μMgcosθ (R) = MR²/2(α)
μ g cosθ = R/2 (α) ...(2)

for pure rolling,
αR = a

μ g cosθ = a / 2
Replace a with the expression in (1)
μ g cosθ = g sinθ - μgcosθ
μ cosθ = sinθ - μ cosθ
2(μ cosθ) = sinθ
μ = tanθ / 2

θ = arc tan (2μ) = arc tan(0.56) = 29.24 deg

3. The Attempt at a Solution
I got incorrect answer for that problem?? I don't know why I got it wrong??

 

[tiny-tim]

Hi Anya91!

μ g cosθ = a / 2
Replace a with the expression in (1)
μ g cosθ = g sinθ - μgcosθ

You dropped the 2.

(btw, you could also have done τ = Iα about the centre of rotation )

 

[Anya91]

thank you so much :) I got correct answer now =)

 

[slk011]

i am so confused.. how can you find the "arc" that you are plugging into "arc tan(0.56) = 29.24 deg" i tried this and i am LOST! helppp!

 

[tiny-tim]

hi slk011!

.. how can you find the "arc" that you are plugging into "arc tan(0.56) = 29.24 deg"

arctan is the same as tan^-1

use the tan^-1 button on your calculator, and it gives you the angle whose tan is 0.56

(you may need to click "2nd" to turn the tan button into tan^-1)

 

[slk011]

Thank you for the help! ... i am using eq (3 tanθ)(tan-1 (3tanθ) and its not working :( what am i doing wrong?

 

[slk011]

nevermind! got it! it was (3 tanθ)+(tan-1 (3tanθ)!