Physics 1 – Solved Practice Problems
(Mechanics & Oscillations)
These problems and solutions were independently
prepared for educational practice purposes.
Question 1: Physics Mechanics: Rotating Disc
Stopping Time Solution
A uniform disc of radius R and mass m is spanned to angular velocity ω and then carefully placed
on a horizontal surface. How long will the disc be rotating on the surface until it comes to a full
stop. Assume that the friction coefficient is μ and the pressure P exerted by the disc on the
surface can be regarded as uniform. Hint: dFf = μdN , dN = P dA, dA = dθrdr, P = πR
mg
2 ,I =
1 2
2 mR
Solution:
Step 1: Normal Force on a Ring Element
Considering a ring of radius r and thickness dr, the normal force is:
2mg
dN = rdr
R2
Step 2: Torque due to Friction
The torque created by friction on this ring is integrated from 0 to R:
R
2μmg 2 2
τ =∫ 2
r dr = μmgR
3
0 R
Step 3: Angular Acceleration
Using the formula:
τ = Iα
where moment of inertia is:
1
I= mR2
2
2
τ μmgR 4μg
α= = 31 =
2 3R
I 2 mR
Final Answer:
Using kinematics
Physics 1 – Solved Practice Problems (Mechanics & Oscillations) 1
, ωf = ωi − αt
with
ωf = 0
ω 3ωR
t= =
4μg
α
Question 2: Pendulum Oscillation with Wall Collision
A ball is suspended by a thread of length l at point O on a wall, forming a small angle $\alpha$ with
the vertical. The string with the ball is deviated by a small angle β (where β > α) and set free. The
solution to the equation of motion for this system is given by θ(t) = β cos(ωt). Assuming the
collision of the ball with the wall is completely elastic, show that the oscillation period of this
pendulum is: T = 2 gl ( π2 + sin−1 ( αβ )). Hint: sin−1 x + cos−1 x = π2
solution:
Step 1: Equation of Motion
The motion of a simple pendulum is given by harmonic motion. Let the release position be at t = 0
with amplitude β .
The equation for angular position θ(t) is:
g
θ(t) = β cos(ωt) where ω=
l
Step 2: Collision Condition
The ball swings from the release point (θ = β ) towards the wall. Since the wall is on the opposite
side, the collision happens at position θ = −α. Let t1 be the time it takes to hit the wall.
α = β cos(ωt1 )
Step 3: Solving for Time (t1 )
We need to isolate t1 . First, divide by β :
α
cos(ωt1 ) = −
β
Using the trigonometric identity cos( π2 + x) = − sin(x), we can find the angle whose cosine is
negative:
+ sin−1 ( )
π α
ωt1 =
2
β
Step 4: Calculating the Period (T )
The oscillation consists of the ball going from β to the wall (−α) and bouncing back to β . Since the
Physics 1 – Solved Practice Problems (Mechanics & Oscillations) 2
(Mechanics & Oscillations)
These problems and solutions were independently
prepared for educational practice purposes.
Question 1: Physics Mechanics: Rotating Disc
Stopping Time Solution
A uniform disc of radius R and mass m is spanned to angular velocity ω and then carefully placed
on a horizontal surface. How long will the disc be rotating on the surface until it comes to a full
stop. Assume that the friction coefficient is μ and the pressure P exerted by the disc on the
surface can be regarded as uniform. Hint: dFf = μdN , dN = P dA, dA = dθrdr, P = πR
mg
2 ,I =
1 2
2 mR
Solution:
Step 1: Normal Force on a Ring Element
Considering a ring of radius r and thickness dr, the normal force is:
2mg
dN = rdr
R2
Step 2: Torque due to Friction
The torque created by friction on this ring is integrated from 0 to R:
R
2μmg 2 2
τ =∫ 2
r dr = μmgR
3
0 R
Step 3: Angular Acceleration
Using the formula:
τ = Iα
where moment of inertia is:
1
I= mR2
2
2
τ μmgR 4μg
α= = 31 =
2 3R
I 2 mR
Final Answer:
Using kinematics
Physics 1 – Solved Practice Problems (Mechanics & Oscillations) 1
, ωf = ωi − αt
with
ωf = 0
ω 3ωR
t= =
4μg
α
Question 2: Pendulum Oscillation with Wall Collision
A ball is suspended by a thread of length l at point O on a wall, forming a small angle $\alpha$ with
the vertical. The string with the ball is deviated by a small angle β (where β > α) and set free. The
solution to the equation of motion for this system is given by θ(t) = β cos(ωt). Assuming the
collision of the ball with the wall is completely elastic, show that the oscillation period of this
pendulum is: T = 2 gl ( π2 + sin−1 ( αβ )). Hint: sin−1 x + cos−1 x = π2
solution:
Step 1: Equation of Motion
The motion of a simple pendulum is given by harmonic motion. Let the release position be at t = 0
with amplitude β .
The equation for angular position θ(t) is:
g
θ(t) = β cos(ωt) where ω=
l
Step 2: Collision Condition
The ball swings from the release point (θ = β ) towards the wall. Since the wall is on the opposite
side, the collision happens at position θ = −α. Let t1 be the time it takes to hit the wall.
α = β cos(ωt1 )
Step 3: Solving for Time (t1 )
We need to isolate t1 . First, divide by β :
α
cos(ωt1 ) = −
β
Using the trigonometric identity cos( π2 + x) = − sin(x), we can find the angle whose cosine is
negative:
+ sin−1 ( )
π α
ωt1 =
2
β
Step 4: Calculating the Period (T )
The oscillation consists of the ball going from β to the wall (−α) and bouncing back to β . Since the
Physics 1 – Solved Practice Problems (Mechanics & Oscillations) 2
