Challenging +1.2 This is a standard center of mass problem for composite bodies requiring the formula for removing mass from a uniform disc. While it involves setting up the geometry correctly (distance OC = 0.5 - r) and solving a quadratic equation, the technique is well-practiced in Further Mechanics courses and follows a predictable method with no novel insight required.
\includegraphics{figure_3}
A circular hole with centre C and radius \(r\) m, where \(r < 0.5\), is cut in a uniform circular disc with centre O and radius 0.5 m. The hole touches the rim of the disc at A (see diagram).
The centre of mass, G, of the remainder of the disc is on the rim of the hole.
Determine the value of \(r\). [5]
Question 3:
3 | (π(0.5)2 −πr2)×2r =...
0.5×π(0.5)2 −r×πr2
r3−0.5r+0.125(=0)
r=0.309 | M1*
A1
A1
M1dep*
A1
[5] | 2.1
3.4
1.1
1.1
2.2a | Forming an equation which if simplified would lead to a three-term
cubic equation in r with no r2term. Do not award this mark until x
has been replaced with a linear expression in r or, if taking moments
about G, zero. Condone absence of πthroughout the question and
for this mark allow inconsistent use of π
Correct LHS (if taking moments about A, C, O or end-point)
Correct RHS – consistent with corresponding LHS
If taking moments about G then award A2 for a correct equation
Re-arranging to obtain a three-term cubic equation/expression in r
only with no r2term (or with πin every term so use of π must now
be consistent) - all terms must be on the same side. The correct
value of r following from a correct equation (in any form) implies
this (and the next) mark – note below some valid method(s) that
may lead to a quadratic equation in r (rather than a cubic)
BC no rationale required for the rejection of the other values of r
−1+ 5
but must be clear that r takes only this value. Accept or
4
awrt 0.309 or allow 0.31 www – note that the correct value of r can
come from the equation r2 +0.5r−0.25=0 (see below)
For reference:
M(A):(π(0.5)2 −πr2)×2r =0.5×π(0.5)2 −r×πr2
M(C):(π(0.5)2 −πr2)×r =(0.5−r)×π(0.5)2 (⇒(0.5−r)(0.5+r)×r =0.25×(0.5−r)∴(0.5+r)×r =0.25)
M(O):(π(0.5)2 −πr2)×(2r−0.5)=(0.5−r)×πr2 (⇒(0.5−r)(0.5+r)×(2r−0.5)=(0.5−r)×r2∴(0.5+r)(2r−0.5)=r2)
M(G):0=(2r−0.5)×π(0.5)2 −r×πr2
M(end-point): (π(0.5)2 −πr2)×(1−2r)=0.5×π(0.5)2 −(1−r)×πr2
\includegraphics{figure_3}
A circular hole with centre C and radius $r$ m, where $r < 0.5$, is cut in a uniform circular disc with centre O and radius 0.5 m. The hole touches the rim of the disc at A (see diagram).
The centre of mass, G, of the remainder of the disc is on the rim of the hole.
Determine the value of $r$. [5]
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2024 Q3 [5]}}