anonymous
  • anonymous
\[\large\mathsf{\text{Problem Based on Rotational Motion}}\]
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
A spool of mass \(\mathsf{m}\) and inner radius \(\mathsf{r}\) and outer radius \(\mathsf{2r}\), having moment of inertia \(\Large\mathsf{\frac{mr^2}{2}}\) is made to roll without sliding on a rough horizontal surface by the help of an applied force \(\mathsf {(F = mg)}\), on ideal string wrapped around the inner cylinder (Shown in the figure). |dw:1326937711360:dw| Find the minimum Co-efficient of Friction required for Pure Rolling.
anonymous
  • anonymous
Here's what I did. |dw:1326938034465:dw| For Pure Rolling, \(\mathsf{\omega . r = v \tag{1}}\) \(\mathsf{\alpha . r = a }\tag{2}\) Two torques are acting on the Spool about it's center (Circle's center), one due to the applied force \(\mathsf F\) and other due to the friction generated by the rolling of the spool on the rough horizontal surface. \[\mathsf{\sum \tau = 2r. f_{k} - F.r = I \alpha }\tag 3\] Translation motion's equation, \[\mathsf{ F - f_{k} = Ma} \tag 4\] Solving equations (2), (3) and (4) and using \(\Large\mathsf{ \mu_{k} = \frac{f_{k}}{N}}\). I am getting \(\Large\mu = \frac{3}{5}\). While the options are \(\large\mathsf{\frac{2}{9}, \frac{4}{9}, \frac{5}{9} \text{and 'none of these'}}\).
anonymous
  • anonymous
Double check a few things. First, note that\[\alpha (2r) = a_{CM}\]Also, note that\[\sum \vec {\bf F}_x = 0\]Otherwise, \[\alpha (2r) = a_{CM}\]won't be satisfied.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Oh yeah, Thank You so much! I knew I did something silly.
anonymous
  • anonymous
Did you get the correct answer?
anonymous
  • anonymous
Yeah, I think so \(\Large\mathsf{\mu_{k} =\frac{5}{9}}\). Thanks!
anonymous
  • anonymous
Excellent.

Looking for something else?

Not the answer you are looking for? Search for more explanations.