Rotational Dynamics - Engineering Mathematics, Lecture notes of Engineering Mathematics

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Dynamics of Rotation

Dynamics of Rotation

Consider only the magnitude :

τ = Fr sinθ , τ is in N-m (MKS) & Dyne-cm (CGS)

Where F – applied force r – moment arm θ – angle between F & r θ

F

Axis Line of Force Moment Arm (r)

Torque (τ)

τ = F x r τ is the Cross Product of the Force vector & Moment Arm

Torque (τ) :

The tendency or ability of a force to rotate an object about an

axis, fulcrum, or pivot. Just as a force is a push or a pull, a

torque can be thought of as a twist.

Torque (τ) :

F 1 F 2 F 3 F 4 r 1 r 2

F

1

r

1

F

2

r

2

1

2

1

= (F

1

)(r

1

) sinθ

1

but θ

2

2

= (F

2

)(r

2

) sinθ

2

2

= (F

2

)(r

2

Torque (Γ) :

F 1 F 2 F 3 F 4 r 1 r 2

F

4

F

3

r

2

3

but θ

3

3

= (F

3

)(r

2

) sinθ

3

3

4

= (F

4

)(0) sinθ

4

4

Dynamics of Rotation

Common Moments of Inertia

http://hyperphysics.phy-astr.gsu.edu I = cMR

c is a fraction

Dynamics of Rotation

Parallel Axes Theorem

I

P

= I

CM

+ Md

Old Axis

• The moment of inertia can be calculated if moved to another axis as long as that (new) axis is parallel to the (original) axis that passes through the center of mass and the distance between them is known. Where: IP : Moment of Inertia about another axis that is parallel to the axis that goes through the center of mass. ICM : Moment of Inertia about an axis at the center of mass. M : Mass of the object d - distance between the center of mass and the point of the parallel axis. d New Axis

A bucket of water with a mass of 20 kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0. 4 m in radius, also with a mass of 20 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20 m to the water. a) What is the tension in the rope while the bucket is falling? b) With what speed does the bucket strike the water? c) What is the time of fall? m = 20 kg R = 0.4 m M = 20 kg h = 20 m T mg a T Mg R R N a = Rα FBD :

Rotation about a FIXED Axis

A bucket of water with a mass of 20 kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0. 4 m in radius, also with a mass of 20 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20 m to the water. a) What is the tension in the rope while the bucket is falling? b) With what speed does the bucket strike the water? c) What is the time of fall? T mg a

NSLM

ΣF

y

= ma ↑+

T – mg =−ma

T =mg −ma >eq- 1

Rotation about a FIXED Axis

A bucket of water with a mass of 20 kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0. 4 m in radius, also with a mass of 20 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20 m to the water. a) What is the tension in the rope while the bucket is falling? b) With what speed does the bucket strike the water? c) What is the time of fall?

−TR = − ½ MRa

T = ½ Ma >eq. 4 eq. 4 = ea. 1

½ Ma = mg – ma

½ Ma + ma = mg

a (½ M+m) = mg

a = g [m/(0.5M + m)]

a = (9.8 m/s

2

) [20 kg /(10 kg + 20 kg )]

a = 6.533 m/s

2

T = ½ (20 kg )(6.533 m/s

2

T = 65.33 N

a in ea. 4

Rotation about a FIXED Axis

A bucket of water with a mass of 20 kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0. 4 m in diameter, also with a mass of 20 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20 m to the water. a) What is the tension in the rope while the bucket is falling? b) With what speed does the bucket strike the water? c) What is the time of fall?

From Kinematics

V

F 2

= V

O 2

+ 2ah h = 20 m V

O

a = 6.533 m/s = 0

2

V

F 2

2

V

F 2

= 261.333 m

2

/s

2

V

F

= 16.167 m/s

Rotation about a FIXED Axis

Work, Energy & Power in Rotational Motion

W = FS, but S = Rθ

W = FRθ , but τ = FR

W = τθ

P

AVE

= W/t

P = (τθ)/t , but ωAVE = θ/t

P AVE = τω AVE

K = ½ MV

2

,but ω = V/R and V = ωR

K = ½ M(ωR)

2

= ½ MR

2

2 ,

but I = MR

2 K = ½ Iω 2 F R R S θ M

Average Power (P^ M

AVE

P (t) = dW/dt

P (t) = τ (dθ/dt) , but ω (t) = dθ/dt

P (t) = τω (t)

Instantaneous Power

t = 20 s I = 245 kg-m 2 R = 4.4 m F = 25 N A playground merry-go-round has a radius of 4. 4 m and a moment of inertia of 245 kg-m 2 and turns with negligible friction about a vertical axle through its center. A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds. a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second interval? b) How much work did the child do on the merry-go-round? c) What is the average power supplied by the child?

Rotation about a FIXED Axis

A playground merry-go-round has a radius of 4. 4 m and a moment of inertia of 245 kg-m 2 and turns with negligible friction about a vertical axle through its center. A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds. a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second interval? b) How much work did the child do on the merry-go-round? c) What is the average power supplied by the child? Work

W = τ θ

τ net = FnetR

= (F – f )R

= FR

Using Torque

τ = (25 N )(4.4 m ) = 110 Nm

θ = ωOt + ½ αt

2 From Kinematics

θ = (0)(20 s ) + ½ (0.45 rad/s

2

)(20 s )

2

θ = 90 rad

W = (110 Nm )(90 rad )

W = 9,900 J = 9.9 KJ

t = 20 s I = 245 kg-m 2 R = 4.4 m F = 25 N

Rotation about a FIXED Axis

A playground merry-go-round has a radius of 4. 4 m and a moment of inertia of 245 kg-m 2 and turns with negligible friction about a vertical axle through its center. A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds. a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second interval? b) How much work did the child do on the merry-go-round? c) What is the average power supplied by the child? From Average Power

PAVE = τ ωAVE

Γ = 110 Nm

PAVE = (110 Nm )(4.5 rad/s )

PAVE = 495 Watts

ωAVE = θ/t

ωAVE = 90 rad /20 s = 4.5 rad/s

θ = 90 rad

t = 20 s I = 245 kg-m 2 R = 4.4 m F = 25 N

Rotation about a FIXED Axis