University Physics Lab: Preliminary Exercises on Rotation and Rolling

This document provides a comprehensive solution to a first-year physics laboratory assignment focusing on the concepts of rotation and rolling. The assignment includes four key problems. The first problem calculates the angular acceleration of a wheel. The second problem addresses the acceleration and angular acceleration of a rolling cylinder on a ramp. The third problem calculates the acceleration of a solid cylinder rolling down a ramp with a given slope, and the fourth problem involves calculating the moment of inertia of a dumbbell with a uniform density. Each problem is solved with detailed explanations, formulas, and step-by-step calculations, including the use of kinematic equations and the moment of inertia formulas for solid cylinders and composite objects. The solution includes references to support the calculations and methodologies used.

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FIRST YEAR PHYSICS LABORATORY PRELIMINARY EXERCISES
EXPERIMENT E2: ROTATION AND ROLLING
Complete the following questions on this sheet, either typed or handwritten, and hand in a hard
copy to a demonstrator at the beginning of your lab session. Show your reasoning and working.
Student’s Name : Number
Group Table Date of exp Mark /15

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In these exercises use g = 9.8 m/s2
1. A wheel accelerates uniformly from rest to a rotational speed of 10 rad/s in 5 s. What
is its angular acceleration?
The angular acceleration of a rotating object is the change of its angular velocity with respect
with the time taken [2]. It is calculated by dividing the angular velocity with the change in time.
Angular acceleration = (change in angular velocity)/ (change in time)
= (final angular velocity- initial angular velocity)/ (final time – initial time)
Where:
α = angular acceleration, (radians/s2)
Δω = change in angular velocity (radians/s)
Δt = change in time (s)
ω1 = initial angular velocity (radians/s)
ω2= final angular velocity (radians/s)
t1 = initial time (s)
t2= final time (s)
from the question,
t1=0, t2=5s, initial angular velocity ω1 = 0 and ω2= 10 radians/sec
therefore, it will be; (10 - 0) / (5 - 0) = 10 ÷ 5 = 2
the angular acceleration α of the wheel is 2 rads / s2
2. A cylinder of diameter 100 mm rolls from rest down a 5 m long ramp and its Centre of
mass is moving with velocity 2 m/s at the bottom of the ramp.
2.1. What is its (constant) acceleration down the ramp?
For this exercise, we apply the kinematic equation. The kinematic equation gives a relationship
between velocities, distance and acceleration of a moving body.
Kinematic equation formula is vf² = vi²
+ 2a d
vi² = initial velocity
a = acceleration
d = displacement
the initial velocity = 0
final velocity Vf2 = 2 x 2 = 4
thus the equation will be vf² = 2a d
working out for a, 4 ÷ 2 x 5= 0.4 m/s2
2.2. What is its angular acceleration?
If it rolls without slipping, the angular acceleration is:
α = a / r
r = ½ (100mm) = 0.05 meters
0.4 / 0.05 = 8 rad/s2

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3. For a solid cylinder, calculate the acceleration as it rolls down a ramp with a slope of
5° above the horizontal
For a rolling body down a ramp,
m a = m g sin (angle)
- I a / R2
(m + I / R2) a = m g sin(angle)
a = m g sin (angle) / (m + I / R2)
m = mass of the solid cylinder, g = acceleration due to gravity, I = moment of inertia and
R = radius
The moment of inertia of a solid cylinder about its central axes is 1/2 m R2
a = m g sin (angle) / (m + 1/2 m) = 2/3 x g sin (angle)
With angle = 50 and g = 9.8 m/s2
a = 2/3 x 9.8 m/s2 x sin (50) =
acceleration = 0.43 m/s2
4. The dumbbell shown below has a uniform density of 2000 kg/m3. Calculate its
moment of inertia about the Centreline.
For a given uniform body, moment of inertia is calculated by finding the moment of inertia of the
middle part of the dumbbell [1].
The moment of inertia of a uniform body is given by; I = (m r2) / 2
The volume of the middle part is Vm = π r2m l, and the mass of the middle part will be the
volume of the middle part x density

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Thus the moment of inertia of the middle part = ½π r4m l ρ
We also need to calculate the volume of the side part. Thus M= Is= π r2s d
The mass will be; πr2s d ρ while the MI will be ½ πr4d ρ
The total MI of the dumbbell is the moment of the middle part plus the moment of the side part;
Thus; ½π r4m l ρ + ½ π r4s d ρ where;
Ρ = 2000 kg/m3
Rm = 0.025 m
L = 0.2 m
Rs = 0.075 m
d = 0.05 m
replacing the values into the equation will be;
(½π x 0.0254 x 0.2 x 2000) + (½π x 0.0754 x 0.05 x 2000)
I = 0.0102 Kg/m2

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References
[1]X. Song, M. Lu and T. Qin, "The Graphics Section Moment of Inertia of an Area Rotation
Axis Calculation Diagram Method——Moment of Inertia of an Area Circle", Advanced
Materials Research, vol. 479-481, pp. 2086-2089, 2012. Available:
10.4028/www.scientific.net/amr.479-481.2086.
[2]"Calculation of angular velocity, angular acceleration and torque of two common point rigid
bodies using IMU", Journal of Applied and Physical Sciences, vol. 3, no. 2, 2017. Available:
10.20474/japs-3.2.3.