OCR MEI Further Mechanics Major 2020 November — Question 12 12 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2020
SessionNovember
Marks12
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Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeParticle in hemispherical bowl
DifficultyChallenging +1.2 This is a standard conical pendulum problem with the added constraint of a hemispherical bowl. Part (a) requires resolving forces in two directions and using circular motion equations—systematic but multi-step. Part (b) is a straightforward deduction from the normal force being positive. While it involves several steps and careful geometry, it follows a well-established template for Further Maths mechanics problems without requiring novel insight.
Spec3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions6.05c Horizontal circles: conical pendulum, banked tracks
\includegraphics{figure_12} Fig. 12 shows a hemispherical bowl. The rim of this bowl is a circle with centre O and radius \(r\). The bowl is fixed with its rim horizontal and uppermost. A particle P, of mass \(m\), is connected by a light inextensible string of length \(l\) to the lowest point A on the bowl and describes a horizontal circle with constant angular speed \(\omega\) on the smooth inner surface of the bowl. The string is taut, and AP makes an angle \(\alpha\) with the vertical.
  1. Show that the normal contact force between P and the bowl is of magnitude \(mg + 2mr\omega^2\cos^2\alpha\). [9]
  2. Deduce that \(g < r\omega^2(k_1 + k_2\cos^2\alpha)\), stating the value of the constants \(k_1\) and \(k_2\). [3]
Question 12:
AnswerMarks Guidance
12(a) r2 =r2 +l2 −2rlcosα⇒l =2rcosα
string
x
sinα= ⇒x=2rcosαsinα
AnswerMarks Guidance
lB1 3.1b
horizontal circle
AnswerMarks Guidance
M1*3.3 Resolving vertically for P – correct
number of terms with components of R
AnswerMarks
and TCondone use of the
same angle for this
mark
AnswerMarks Guidance
Rsinθ=Tcosα+mgA1 1.1
string, R is the normal
contact force and θ is
the angle between the
horizontal and the
normal contact force
AnswerMarks Guidance
M1*3.3 NII horizontally – correct number of
terms – allow any form for radiusCondone use of the
same angle
AnswerMarks Guidance
Rcosθ+Tsinα=m(2rcosαsinα)ω2A1 1.1
2rcosαsinα
cosθ= =2cosαsinα or
r
r−2rcos2α
sinθ= =1−2cos2α
AnswerMarks Guidance
rB1 1.1
sinθ in terms of αor stating a correct
relationship between θand αe.g.
θ=2α−1π
2
( )
R 1−2cos2α =Tcosα+mg
AnswerMarks Guidance
R(2cosαsinα)+Tsinα=2mrω2cosαsinαM1dep* 3.4
previous M marks
T =2mrω2cosα−2Rcosα
( ) ( )
⇒R 1−2cos2α = 2mrω2cosα−2Rcosα cosα+mg
R−2Rcos2α=2mrω2cos2α−2Rcos2α+mg
AnswerMarks Guidance
R=mg+2mrω2cos2αA1 2.2a
[9]
AnswerMarks Guidance
12(b) ( )
T =2mcosα rω2 −g−2rω2cos2αM1* 3.4
expression for T in terms of r,m,g,α
and ω
AnswerMarks Guidance
T >0⇒rω2 −g−2rω2cos2α>0M1dep* 3.4
( )
AnswerMarks Guidance
g <rω2 1−2cos2αA1 2.2a
1 2
[3]
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Question 12:
12 | (a) | r2 =r2 +l2 −2rlcosα⇒l =2rcosα | B1 | 3.1b | l is the length of the
string
x
sinα= ⇒x=2rcosαsinα
l | B1 | 3.1b | oe | x is the radius of the
horizontal circle
M1* | 3.3 | Resolving vertically for P – correct
number of terms with components of R
and T | Condone use of the
same angle for this
mark
Rsinθ=Tcosα+mg | A1 | 1.1 | T is the tension in the
string, R is the normal
contact force and θ is
the angle between the
horizontal and the
normal contact force
M1* | 3.3 | NII horizontally – correct number of
terms – allow any form for radius | Condone use of the
same angle
Rcosθ+Tsinα=m(2rcosαsinα)ω2 | A1 | 1.1 | oe
2rcosαsinα
cosθ= =2cosαsinα or
r
r−2rcos2α
sinθ= =1−2cos2α
r | B1 | 1.1 | Either correct expression for cosθ or
sinθ in terms of αor stating a correct
relationship between θand αe.g.
θ=2α−1π
2
( )
R 1−2cos2α =Tcosα+mg
R(2cosαsinα)+Tsinα=2mrω2cosαsinα | M1dep* | 3.4 | Eliminate θ from both equations | Dependent on both
previous M marks
T =2mrω2cosα−2Rcosα
( ) ( )
⇒R 1−2cos2α = 2mrω2cosα−2Rcosα cosα+mg
R−2Rcos2α=2mrω2cos2α−2Rcos2α+mg
R=mg+2mrω2cos2α | A1 | 2.2a | AG
[9]
12 | (b) | ( )
T =2mcosα rω2 −g−2rω2cos2α | M1* | 3.4 | Use given expression for R to find
expression for T in terms of r,m,g,α
and ω
T >0⇒rω2 −g−2rω2cos2α>0 | M1dep* | 3.4 | Setting T > 0
( )
g <rω2 1−2cos2α | A1 | 2.2a | k =1 and k =−2
1 2
[3]
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
\includegraphics{figure_12}

Fig. 12 shows a hemispherical bowl. The rim of this bowl is a circle with centre O and radius $r$. The bowl is fixed with its rim horizontal and uppermost.

A particle P, of mass $m$, is connected by a light inextensible string of length $l$ to the lowest point A on the bowl and describes a horizontal circle with constant angular speed $\omega$ on the smooth inner surface of the bowl.

The string is taut, and AP makes an angle $\alpha$ with the vertical.

\begin{enumerate}[label=(\alph*)]
\item Show that the normal contact force between P and the bowl is of magnitude $mg + 2mr\omega^2\cos^2\alpha$. [9]
\item Deduce that $g < r\omega^2(k_1 + k_2\cos^2\alpha)$, stating the value of the constants $k_1$ and $k_2$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2020 Q12 [12]}}
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