Circular and Rotational Motion Practice Flashcards

0.0(0)
Studied by 0 people
call kaiCall Kai
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/31

flashcard set

Earn XP

Description and Tags

Vocabulary-style flashcards based on lecture notes covering Kinematics, Dynamics, Rotational Motion, and Rolling Motion formulas and concepts.

Last updated 6:39 AM on 6/1/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai

No analytics yet

Send a link to your students to track their progress

32 Terms

1
New cards

Angular Velocity (ω\omega)

The rate of change of angular displacement, given by ω=dθdt=2πn=2πT\omega = \frac{d\theta}{dt} = 2\pi n = \frac{2\pi}{T}, where nn is frequency and TT is the time period.

2
New cards

Relation between Linear and Angular Velocity

v=rωv = r\omega (Vector form: v=ω×r\mathbf{v} = \mathbf{\omega} \times \mathbf{r}).

3
New cards

Centripetal Acceleration (aca_c)

The acceleration directed toward the center of a circular path, expressed as ac=v2r=rω2=vωa_c = \frac{v^2}{r} = r\omega^2 = v\omega.

4
New cards

Centripetal Force (FcF_c)

Fc=mv2r=mrω2=mvωF_c = \frac{mv^2}{r} = mr\omega^2 = mv\omega (Directed towards the center).

5
New cards

Max Safe Speed on Unbanked Road (vmaxv_{max})

vmax=μsrgv_{max} = \sqrt{\mu_s rg}, where μs\mu_s is the coefficient of static friction.

6
New cards

Angle of Banking (θ\theta)

θ=tan1(v2rg)\theta = \tan^{-1}\left( \frac{v^2}{rg} \right).

7
New cards

Optimum Speed (v0v_0)

The ideal speed for a banked road to minimize wear on tires, given by v0=rgtan(θ)v_0 = \sqrt{rg \tan(\theta)}.

8
New cards

Max Safe Speed on Banked Road (with friction)

vmax=rg(μs+tan(θ)1μstan(θ))v_{max} = \sqrt{rg \left( \frac{\mu_s + \tan(\theta)}{1 - \mu_s \tan(\theta)} \right)}.

9
New cards

Min Safe Speed on Banked Road (with friction)

vmin=rg(tan(θ)μs1+μstan(θ))v_{min} = \sqrt{rg \left( \frac{\tan(\theta) - \mu_s}{1 + \mu_s \tan(\theta)} \right)}.

10
New cards

Conical Pendulum Time Period (TT)

T=2πLcos(θ)g=2πhgT = 2\pi \sqrt{\frac{L \cos(\theta)}{g}} = 2\pi \sqrt{\frac{h}{g}}, where LL is the length and hh is the vertical height.

11
New cards

VCM Top Position (Highest)

Minimum velocity vmin=rgv_{min} = \sqrt{rg} and Tension Ttop0T_{top} \geq 0.

12
New cards

VCM Bottom Position (Lowest)

Minimum velocity vmin=5rgv_{min} = \sqrt{5rg} and Tension Tbottom=6mgT_{bottom} = 6mg.

13
New cards

VCM Midway Position (Horizontal)

Minimum velocity vmin=3rgv_{min} = \sqrt{3rg} and Tension Tmid=3mgT_{mid} = 3mg.

14
New cards

Moment of Inertia (M.I.) (II)

The sum of the products of the mass of each particle and the square of its distance from the axis of rotation: I=miri2=r2dmI = \sum m_i r_i^2 = \int r^2 dm.

15
New cards

Radius of Gyration (kk)

k=IMk = \sqrt{\frac{I}{M}}; the distance from the axis where the entire mass can be assumed to be concentrated.

16
New cards

Parallel Axes Theorem

I0=Ic+Mh2I_0 = I_c + Mh^2, where IcI_c is the moment of inertia through the center of mass and hh is the distance between axes.

17
New cards

Perpendicular Axes Theorem

Iz=Ix+IyI_z = I_x + I_y (Applicable for planar bodies only).

18
New cards

M.I. of a Ring (IcI_c)

MR2MR^2 (Axis through center, perpendicular to plane).

19
New cards

M.I. of a Disc (IcI_c)

MR2MR^2.

20
New cards

M.I. of a Solid Cylinder (IcI_c)

MR2MR^2 (Same as disc).

21
New cards

M.I. of a Hollow Cylinder (IcI_c)

MR2MR^2 (Same as ring).

22
New cards

M.I. of a Solid Sphere (IcI_c)

MR2MR^2.

23
New cards

M.I. of a Hollow Sphere / Shell (IcI_c)

MR2MR^2.

24
New cards

M.I. of a Thin Rod (IcI_c)

112ML2\frac{1}{12} ML^2 (Axis through center, perpendicular to length).

25
New cards

Torque (τ\tau)

τ=Iα\tau = I\alpha (Vector form: τ=r×F\mathbf{\tau} = \mathbf{r} \times \mathbf{F}).

26
New cards

Angular Momentum (LL)

L=IωL = I\omega (Vector form: L=r×p\mathbf{L} = \mathbf{r} \times \mathbf{p}).

27
New cards

Law of Conservation of Angular Momentum

If net external torque τext=0\tau_{ext} = 0, then L=constantL = \text{constant}, or I1ω1=I2ω2I_1\omega_1 = I_2\omega_2.

28
New cards

Rotational Kinetic Energy (ErE_r)

Er=12Iω2=L22IE_r = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}.

29
New cards

Work Done in Rotation (WW)

W=τθW = \tau \theta.

30
New cards

Total Kinetic Energy (ETE_T) in Rolling

The sum of translational and rotational kinetic energy: ET=Et+Er=12mv2+12Iω2E_T = E_t + E_r = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2.

31
New cards

Acceleration Area of a body rolling down an Inclined Plane

a=gsin(θ)1+k2R2a = \frac{g \sin(\theta)}{1 + \frac{k^2}{R^2}}, where θ\theta is the angle of the incline.

32
New cards

Velocity at bottom of an Inclined Plane

v=2gh1+k2R2v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}.