A wheel starts from rest and accelerates uniformly at α = 2.0 rad/s² for 5.0 seconds. Find: (a) the final angular velocity, (b) the total angle rotated, and (c) the number of complete revolutions.
Angular velocity, acceleration, and torque with calculus
How can I study Rotational Kinematics and Dynamics effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Rotational Kinematics and Dynamics study guide free?▾
Yes — all study notes, flashcards, and practice problems for Rotational Kinematics and Dynamics on Study Mondo are 100% free. No account is needed to access the content.
What course covers Rotational Kinematics and Dynamics?▾
Rotational Kinematics and Dynamics is part of the AP Physics C: Mechanics course on Study Mondo, specifically in the Rotational Motion section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Rotational Kinematics and Dynamics?
θ
Angular acceleration:α=dtdω=dt2d2θ
Relationship to Linear Quantities
For point at distance r from axis:
Arc length:s=rθ
Linear velocity:v=rω
Tangential acceleration:at=rα
Centripetal acceleration:ac=rv2=rω2
Rotational Kinematics
For constant angular acceleration α:
ω=ω0+αt
θ=θ0+ω0t+21αt2
ω2=ω02+2α(θ−θ0)
(Analogous to linear kinematics)
Variable Angular Acceleration
When α(t) is given:
ω(t)=ω0+∫0tα(t′)dt′
θ(t)=θ0+∫0tω(t′)dt′
When α(θ) is given, use:
α=dtdω=dθdωdtdθ=ωdθdω
Total kinetic energy:KE=KEtrans+KErot=21Mvcm2+21Icmω2
Rolling without slipping:vcm=Rω
where R is radius.
Example: Rolling Down Incline
Energy conservation:
Mgh=21Mv2+21Iω2
Using v=Rω and I=βMR2:
Mgh=21Mv2+21βMR2R2v
gh=21v2(1+β)
v=1+β2gh
where β depends on shape:
Solid cylinder: β=1/2
Solid sphere: β=2/5
Hoop: β=1
Angular Impulse
∫t1t2τdt=ΔL=IΔω
where L=Iω is angular momentum.
ω=ω0+αt=0+(2.0)(5.0)
ω=10 rad/s
(b) Total angle rotated:
θ=ω0t+21αt2=0+21(2.0)(5.0)2
θ=(1.0)(25)
θ=25 rad
(c) Number of revolutions:
N=2πθ=2π25
N=3.98 rev≈4 revolutions
2Problem 2medium
❓ Question:
A solid disk (mass M = 5.0 kg, radius R = 0.3 m) rotates about its central axis. A tangential force F = 15 N is applied at the rim. Find: (a) the torque, (b) the angular acceleration, and (c) the angular velocity after 3.0 seconds (starting from rest).
💡 Show Solution
Given:
M = 5.0 kg
R = 0.3 m
F = 15 N (tangential at rim)
ω₀ = 0
(a) Torque:
τ=rF=RF=(0.3)(15)
τ=4.5 N\cdotpm
(b) Angular acceleration:
Moment of inertia for solid disk:
I=21MR
α=Iτ=0.225
α=20 rad/s2
(c) Angular velocity after 3.0 s:
ω=ω0+αt=0+(20)(3.0)
ω=60 rad/s
Linear speed at rim:
v=Rω=(0.3)(60)=18 m/s
3Problem 3hard
❓ Question:
A uniform rod (length L = 1.0 m, mass M = 2.0 kg) is hinged at one end and held horizontal. It is released from rest. Find: (a) the initial angular acceleration, (b) the angular velocity when the rod is vertical, and (c) the linear speed of the free end when vertical.
💡 Show Solution
Given:
L = 1.0 m
M = 2.0 kg
Rod hinged at one end, released from horizontal
(a) Initial angular acceleration:
Moment of inertia about end:
I=31ML2=31
Torque due to gravity (acts at CM, distance L/2 from hinge):
τ=Mg2L=(2.0)(9.8)
α0=Iτ=
α0=14.7 rad/s2
(b) Angular velocity when vertical:
Using energy conservation:
Initial PE (relative to final): Ui=Mg2L (CM drops by L/2)
Final KE: Kf=21Iω
Mg2L=21
ω2=IMgL=
ω2=29.4
ω=5.42 rad/s
(c) Linear speed of free end:
v=Lω=(1.0)(5.42)
v=5.42 m/s
Check with energy:
21Mvcm2=
vcm=gL
Since vend=2vcm for rotation about end: m/s ✓
▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.