OCR MEI Further Mechanics A AS 2024 June — Question 2 11 marks

Exam BoardOCR MEI
ModuleFurther Mechanics A AS (Further Mechanics A AS)
Year2024
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeDimensional analysis MI context
DifficultyModerate -0.8 This is a straightforward dimensional analysis question with standard bookwork parts (a-c) requiring only recall of basic dimensions and simple algebraic manipulation in part (d). Part (e) is simple substitution. All steps are routine with no novel problem-solving required, making it easier than average.
Spec6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions6.01e Formulate models: dimensional arguments

2
  1. Find the dimensions of energy. The moment of inertia, \(I\), of a rigid body rotating about a fixed axis is measured in \(\mathrm { kg } \mathrm { m } ^ { 2 }\).
  2. State the dimensions of \(I\). The kinetic energy, \(E\), of a rigid body rotating about a fixed axis is given by the formula \(\mathrm { E } = \frac { 1 } { 2 } \mathrm { I } \omega ^ { 2 }\),
    where \(\omega\) is the angular velocity (angle per unit time) of the rigid body.
  3. Show that the formula for \(E\) is dimensionally consistent. When a rigid body is pivoted from one of its end points and allowed to swing freely, it forms a pendulum. The period, \(t\), of the pendulum is the time taken for it to complete one oscillation. A student conjectures the formula \(\mathrm { t } = \left. \mathrm { k } ( \mathrm { mg } ) ^ { \alpha } \mathrm { r } ^ { \beta } \right| ^ { \gamma }\),
    where
    The moment of inertia of a thin uniform rigid rod of mass 1.5 kg and length 0.8 m , rotating about one of its endpoints, is \(0.32 \mathrm {~kg} \mathrm {~m} ^ { 2 }\). The student suspends such a rod from one of its endpoints and allows it to swing freely. The student measures the period of this pendulum and finds that it is 1.47 seconds.
  4. Using the formula conjectured by the student, determine the value of \(k\).

Question 2:
AnswerMarks Guidance
2(a) e.g.  E  =  12 m v 2  = M ( L T − 1 ) 2 = M L 2 T − 2
[1]1.1 Could be from any known expression which
calculates work done or an energy (like GPE, EPE)
Correct answer www earns B1. B0 for kg m2s−2
AnswerMarks Guidance
(b)ML2 B1
[1]1.1
(c)  T 1  = −
So 12 I 2 M L 2 T 2  E     =  − = (as required)M1
A1
AnswerMarks
[2]1.2
1.1Stated or clearly implied
e.g. [𝜔2] = (T−1)2 or [𝜔2] = T−2
Just ML2T−2 = ML2(T−1)2 is M0, but this earns
M1 when accompanied by 𝐸 = 1𝐼𝜔2
2
[RHS] = ML2(T−1)2 earns M1
Conclusion stated (may be just a tick) after correct
working
AnswerMarks
(d)T=(MLT −2)  L  (ML2) 
1
[ 2 1  − = , so ] 𝛼 = −
2
0   + =
++2=0
=−1,= 1
AnswerMarks
2 2M1
B1
M1
A1
AnswerMarks
[4]1.1
1.1
1.1a
AnswerMarks
1.1Substituting dimensional expressions into formula;
accept their expression for  I  ; may already be
expanded. Allow use of units kg, m, s
Allow one error in [mg ] and/or [ r ]
Equating indices of M and L Allow one error (but
not omission).
Three correct equations imply first M1
Correct values www imply M1A1
Three correct values www imply M1B1M1A1
AnswerMarks
(e)1
𝑟 = ×0.8 (= 0.4)
2
0 .3 2
1 .4 7 = k
(1 .5 ) ( 9 .8 )  12 ( 0 .8 )
AnswerMarks
 k = 6 .3 0 (1 3 1 )B1
M1
A1
AnswerMarks
[3]3.1b
3.4
AnswerMarks
1.1Substituting data FT indices (at least two non-zero)
Allow 𝑟 = 0.8 and one other error
A1 FT for answer of 8.91 (when 𝑟 = 0.8 used)
Otherwise cao
Question 2:
2 | (a) | e.g.  E  =  12 m v 2  = M ( L T − 1 ) 2 = M L 2 T − 2 | B1
[1] | 1.1 | Could be from any known expression which
calculates work done or an energy (like GPE, EPE)
Correct answer www earns B1. B0 for kg m2s−2
(b) | ML2 | B1
[1] | 1.1
(c) |   T 1  = −
So 12 I 2 M L 2 T 2  E     =  − = (as required) | M1
A1
[2] | 1.2
1.1 | Stated or clearly implied
e.g. [𝜔2] = (T−1)2 or [𝜔2] = T−2
Just ML2T−2 = ML2(T−1)2 is M0, but this earns
M1 when accompanied by 𝐸 = 1𝐼𝜔2
2
[RHS] = ML2(T−1)2 earns M1
Conclusion stated (may be just a tick) after correct
working
(d) | T=(MLT −2)  L  (ML2) 
1
[ 2 1  − = , so ] 𝛼 = −
2
0   + =
++2=0
=−1,= 1
2 2 | M1
B1
M1
A1
[4] | 1.1
1.1
1.1a
1.1 | Substituting dimensional expressions into formula;
accept their expression for  I  ; may already be
expanded. Allow use of units kg, m, s
Allow one error in [mg ] and/or [ r ]
Equating indices of M and L Allow one error (but
not omission).
Three correct equations imply first M1
Correct values www imply M1A1
Three correct values www imply M1B1M1A1
(e) | 1
𝑟 = ×0.8 (= 0.4)
2
0 .3 2
1 .4 7 = k
(1 .5 ) ( 9 .8 )  12 ( 0 .8 )
 k = 6 .3 0 (1 3 1 ) | B1
M1
A1
[3] | 3.1b
3.4
1.1 | Substituting data FT indices (at least two non-zero)
Allow 𝑟 = 0.8 and one other error
A1 FT for answer of 8.91 (when 𝑟 = 0.8 used)
Otherwise cao
2
\begin{enumerate}[label=(\alph*)]
\item Find the dimensions of energy.

The moment of inertia, $I$, of a rigid body rotating about a fixed axis is measured in $\mathrm { kg } \mathrm { m } ^ { 2 }$.
\item State the dimensions of $I$.

The kinetic energy, $E$, of a rigid body rotating about a fixed axis is given by the formula\\
$\mathrm { E } = \frac { 1 } { 2 } \mathrm { I } \omega ^ { 2 }$,\\
where $\omega$ is the angular velocity (angle per unit time) of the rigid body.
\item Show that the formula for $E$ is dimensionally consistent.

When a rigid body is pivoted from one of its end points and allowed to swing freely, it forms a pendulum. The period, $t$, of the pendulum is the time taken for it to complete one oscillation. A student conjectures the formula\\
$\mathrm { t } = \left. \mathrm { k } ( \mathrm { mg } ) ^ { \alpha } \mathrm { r } ^ { \beta } \right| ^ { \gamma }$,\\
where

\begin{itemize}
  \item $k$ is a dimensionless constant,
  \item $m$ is the mass of the rigid body,
  \item $g$ is the acceleration due to gravity,
  \item $r$ is the distance between the pivot point and the rigid body's centre of mass.
\item Use dimensional analysis to find the values of $\alpha , \beta$ and $\gamma$.
\end{itemize}

The moment of inertia of a thin uniform rigid rod of mass 1.5 kg and length 0.8 m , rotating about one of its endpoints, is $0.32 \mathrm {~kg} \mathrm {~m} ^ { 2 }$. The student suspends such a rod from one of its endpoints and allows it to swing freely. The student measures the period of this pendulum and finds that it is 1.47 seconds.
\item Using the formula conjectured by the student, determine the value of $k$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2024 Q2 [11]}}