Physics 111 Torque and Moments of Inertia Equations

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Physics 111 Torque and Moments of Inertia

Physics 111 Torque and Moments of Inertia

Equations: �Torque (t) = F * L ◦ T=F * d sin ɵ If

Equations: �Torque (t) = F * L ◦ T=F * d sin ɵ If not rotation. Delta Fy: 0 Delta Fx: 0 Torque : 0 For uniform distribution of mass Weight is @. 5 L Summation of t = I ᾰ I = Summation of mr^2

Problem 1 �The pull cord of a lawnmower engine is wound around a drum

Problem 1 �The pull cord of a lawnmower engine is wound around a drum of radius 6 00 cm while the cord is pulled with a force of 75. 0 N to start the engine. What magnitude torque does the cord apply to the drum?

Solution: �T = r*f �(. 06 * 75) = 4. 5 N * m

Solution: �T = r*f �(. 06 * 75) = 4. 5 N * m

Practice Problem 2: �Revolutionaries attempt to pull down a statue of the Great Leader

Practice Problem 2: �Revolutionaries attempt to pull down a statue of the Great Leader by pulling on a rope tied to the top of his head. The statue is 17 m tall, and they pull with a force of 4200 N at an angle of 65° to the horizontal. �What is the torque they exert on the statue? �If they are standing to the right of the statue, is the torque positive or negative?

Solution: �Given: ◦ height= 17 m ◦ F = 4200 ◦ Angle = 65

Solution: �Given: ◦ height= 17 m ◦ F = 4200 ◦ Angle = 65 Find: torque First find the length between the axis of rotation and the line of action sin(65) = 17 m / L L = 17 m / sin 45 = 20 m T = l * F = 20 * 4200 = 83910 N. m Is positive, so moving ccw

Practice Problem: �What is the rotational inertia of a solid iron disk of mass

Practice Problem: �What is the rotational inertia of a solid iron disk of mass 49 0 kg with a thickness of 5 00 cm solid iron disk of mass and a radius of 20. 0 cm, about an axis through its center and perpendicular to it? �Need to use moments of inertia of common shapes table!

Solution: �For disk I =. 5 mr^2 �Need to convert cm to meters. �I

Solution: �For disk I =. 5 mr^2 �Need to convert cm to meters. �I =. 5 ( 49)(. 2)^2 =. 98 kg * m^2

Problem: Find the moment of inertia of the � system below. The masses are

Problem: Find the moment of inertia of the � system below. The masses are m 1 an d m 2 and they are separated by a distance r. Assume the rod connecting the masses is massless. ( moving ccw) �Take m 1 = 2. 00 kg, m 2 = 1. 00 kg, r 1= 0. 33 m , and r 2 = 0. 67 m. M 1 M 2

Solution: �I = summation of mr^2 �I = m 1 r 1^2 + m

Solution: �I = summation of mr^2 �I = m 1 r 1^2 + m 2 r 2^2 �= (2 kg)(. 33 m)^2 + (1 kg)(. 67 m)^2 �=. 67 kg m^2

Part 2 : �What is the moment of inertia if the axis is moved

Part 2 : �What is the moment of inertia if the axis is moved so that is passes through m 1? �What does this mean?

Solution Part 2: �I = summation of mr^2 �I = m 1 r 1^2

Solution Part 2: �I = summation of mr^2 �I = m 1 r 1^2 + m 2 r 2^2 �= (2 kg)(0)^2 + (1 kg)(1 m)^2 �= 1 kg m^2 �Moment of Inertia increased!

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