backward from the cylinder. The cylinder experiences a forward friction force from the plank. : For the plank: For the cylinder (linear): For the cylinder (rotational about center of mass):
2θ̇θ̈=−3gLcosθ⋅θ̇2 theta dot theta double dot equals negative the fraction with numerator 3 g and denominator cap L end-fraction cosine theta center dot theta dot
Below, we have curated a comprehensive list of the best resources—from classic textbooks to online treasure troves—to help you master the mechanics of your next physics competition.
If you are looking to elevate your physics game and access hundreds of curated problems like these, visit our master directory. backward from the cylinder
Because the string is fixed to the ceiling and does not slip on the cylinder, the kinematic constraint linking translation and rotation is: a=Rα(Equation 3)a equals cap R alpha space (Equation 3) Substitute Equation 3 into Equation 2: T=12Macap T equals one-half cap M a Now, substitute this expression for into Equation 1: Mg−12Ma=Macap M g minus one-half cap M a equals cap M a
, the derivative is positive, meaning the equilibrium is .
ax,cm=ddt(−L2sinθ⋅θ̇)=−L2(cosθ⋅θ̇2+sinθ⋅θ̈)a sub x comma c m end-sub equals d over d t end-fraction open paren negative the fraction with numerator cap L and denominator 2 end-fraction sine theta center dot theta dot close paren equals negative the fraction with numerator cap L and denominator 2 end-fraction open paren cosine theta center dot theta dot squared plus sine theta center dot theta double dot close paren If you are looking to elevate your physics
K=12M(L24θ̇2)+12(112ML2)θ̇2=16ML2θ̇2cap K equals one-half cap M open paren the fraction with numerator cap L squared and denominator 4 end-fraction theta dot squared close paren plus one-half open paren 1 over 12 end-fraction cap M cap L squared close paren theta dot squared equals one-sixth cap M cap L squared theta dot squared By Conservation of Energy (
: Caltech's high-quality, open-access repository of fundamental physics derivations, ideal for building core intuition for olympiad-level mechanics.
Iputty=M(L4)2=116ML2=348ML2cap I sub p u t t y end-sub equals cap M open paren the fraction with numerator cap L and denominator 4 end-fraction close paren squared equals 1 over 16 end-fraction cap M cap L squared equals 3 over 48 end-fraction cap M cap L squared Total moment of inertia: Iputty=M(L4)2=116ML2=348ML2cap I sub p u t t y
This public link is valid for 7 days and shares a thread, including any personal information you added. This link or copies made by others cannot be deleted. If you share with third parties, their policies apply. Can’t copy the link right now. Try again later.
The combined system rotates as a rigid body with angular velocity about the CoM. We must find the new total moment of inertia Itotalcap I sub t o t a l end-sub about this pivot. Using the Parallel Axis Theorem for the rod: