Sphere rolling down an incline

AI Thread Summary
A hollow spherical shell with a mass of 2.50 kg rolls down a slope at a 32-degree angle, leading to calculations of its acceleration, frictional force, and the minimum coefficient of friction required to prevent slipping. The acceleration of the center of mass is calculated as 3.12 m/s² using the formula a_c = (3/5)g sin(θ). The frictional force acting on the shell is determined to be 5.19 N. To find the minimum coefficient of friction, it is suggested that the frictional force is at its maximum, leading to a calculated value of μ = 0.246. The discussion emphasizes the relationship between rolling motion and friction in determining these values.
dk214
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Homework Statement


A hollow spherical shell with mass 2.50 kg rolls without slipping down a slope that makes an angle of 32.0 degrees with the horizontal.
a. Find the magnitude of the acceleration a_c of the center of mass of the spherical shell.
b. Find the magnitude of the frictional force acting on the spherical shell.
c. Find the minimum coefficient of friction \mu needed to prevent the spherical shell from slipping as it rolls down the slope.


Homework Equations


For part a.
Since its pure roll, a_c = \alpha * R <br /> \alpha = a_c/R
\tau = R*Friction = I (moment-of-inertia) * \alpha
Friction = (I*\alpha)/R = (I*a_c)/R^2
Ma_c = Mgsin(\theta)-Friction
Ma_c = Mgsin(\theta)-Ia_c/R^2
a_c = (MR^2*g*sin(\theta))/(MR^2+I)


The Attempt at a Solution


I for sphere =2/3 MR^2
so, a_c = (MR^2*g*sin(\theta))/(MR^2+2/3*MR^2)
MR^2 cancels..
a_c = 3/5*g*sin(\theta)
for a_c i got a_c = 3.12m/s^2 i think I am right unless i made a mathematical error some where.
and substituting a_c, in Ma_c = Mgsin(\theta)-Friction
i got Friction = 5.19 N.
And c,
this where I'm kind of stuck. I'm assuming since they are asking for minimum \mu Friction is 0 in Ma_c = Mgsin(\theta)-Friction [\tex]<br /> a_c = gsin(\theta). <br /> Friction = (I*\alpha)/R = (I*a_c)/R^2, and <br /> Friction = \mu*mg*sin(\theta) <br /> /mu= ((I*a_c)/R^2)/mg*sin(\theta)&lt;br /&gt; idk if I&amp;#039;m right in assuming Friction is 0 in one part and not in other.. Any hints/guides and help would greatly be appreciated.
 
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Hi dk214,

dk214 said:

The Attempt at a Solution


I for sphere =2/3 MR^2
so, a_c = (MR^2*g*sin(\theta))/(MR^2+2/3*MR^2)
MR^2 cancels..
a_c = 3/5*g*sin(\theta)
for a_c i got a_c = 3.12m/s^2 i think I am right unless i made a mathematical error some where.
and substituting a_c, in Ma_c = Mgsin(\theta)-Friction
i got Friction = 5.19 N.
And c,
this where I'm kind of stuck. I'm assuming since they are asking for minimum \mu Friction is 0 in

I don't think this is right; you've already found the force of friction. Now they want the minimum \mu that can supply that force; in other words they want the coefficient for which that frictional force is a maximum. What does that give?
 
I don't know if I'm understanding the question right. Are they just asking for the \mu for the friction I found.?
which would just be Friction/Normal
\mu = 5.12/Mgcos(\theta)
\mu = .246
 
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