Lesson 4.3 Problem Solving in Rotation

Lesson 4.3 Problem Solving in Rotation - Problem solving in...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem solving in Rotational Motion In order to solve problems in rotational motion you need to recall all the formulae we developed earlier. It is easier to write down the formulae for linear motion and then write the corresponding equation for rotational motion. Linear motion Rotational motion F = m a τ = I α W = F x P = m v L = I ϖ 2 2 1 1 2 2 rolling cm K I mv ϖ = + 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
A 3.5 kg cylinder of radius 15 cm is initially at rest. A rope of negligible mass is wrapped around it and pulled with a force of 17 N. Find (a) the torque exerted by the rope, (b) the angular acceleration of the cylinder and (c) the angular velocity of the cylinder after 5.0 s. m = 3.5 kg, r = 15 cm = 0.15 m, F = 17 N (a) F = 17 N is at right angles to r = 0.15 m τ = r . F sin θ = r. F = 0.15 × 17 = 2.55 mN (a) To find α , we need the moment of inertia I of the cylinder. A cylinder is the same as a disc of radius r Moment of inertia of a cylinder is τ = I α F = 17 N I τ α= 2 2.55 m.N 0.04 kg.m = -2 63.75 rad.s = 0.15 m 2
Background image of page 2
(c) the angular velocity of the cylinder after 5.0 s. F = 17 N = 0 + 63.75 × 5 = 318.75 rad s -1 0.15 m 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
A wheel mounted on an axle that is not frictionless is initially at rest. A constant external torque of 50 N is applied to the wheel for 20 s giving it an angular velocity of 600 rpm. The external torque is then removed and the wheel comes to rest in 120 s. Find (a) the moment of inertia of the wheel, (b) the frictional torque that brings the wheel to a stop. We first find the angular acceleration α Now we use the relation τ = I α and solve for I I τ α = f o t ϖ - = -2 50 m.N rad.s π = -1 20 rad.s 20 s = -2 rad.s = 2 50 kg.m = τ = 50 m.N, t = 20 s 4
Background image of page 4
(a) When the external torque is removed the frictional torque τ f produces a negative angular acceleration bringing the wheel to a stop f τ = Iα f o t ϖ α - = -1 0 20 rad.s 120 s π - = -2 rad.s 6 = - 2 -2 50 = kg.m rad.s 6 3 m N = - 5
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 21

Lesson 4.3 Problem Solving in Rotation - Problem solving in...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online