Sphere rolling down an incline

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SUMMARY

A hollow spherical shell with a mass of 2.50 kg rolls down a slope at an angle of 32.0 degrees, resulting in a center of mass acceleration of 3.12 m/s². The frictional force acting on the shell is calculated to be 5.19 N. To prevent slipping, the minimum coefficient of friction (\mu) required is determined to be 0.246. These calculations utilize the moment of inertia for a sphere, I = 2/3 MR², and the relationship between linear and angular acceleration.

PREREQUISITES
  • Understanding of rotational dynamics and moment of inertia
  • Familiarity with Newton's second law for rotational motion
  • Knowledge of the relationship between linear and angular acceleration
  • Basic trigonometry, particularly sine functions in relation to angles
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes, focusing on hollow spheres
  • Learn about the dynamics of rolling motion and the conditions for rolling without slipping
  • Explore the effects of friction on motion, particularly in inclined planes
  • Investigate the application of Newton's laws in both linear and rotational contexts
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of rolling motion and friction in inclined planes.

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 [tex]a_c[/tex] 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 [tex]\mu[/tex] needed to prevent the spherical shell from slipping as it rolls down the slope.


Homework Equations


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


The Attempt at a Solution


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

dk214 said:

The Attempt at a Solution


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

I don't think this is right; you've already found the force of friction. Now they want the minimum [itex]\mu[/itex] 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 [tex]\mu[/tex] for the friction I found.?
which would just be Friction/Normal
[tex]\mu = 5.12/Mgcos(\theta)[/tex]
[tex]\mu = .246[/tex]
 

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