Download Page 1 PES 1110 Fall 2013, Spendier Lecture 31/Page 1 Today

PES 1110 Fall 2013, Spendier
Lecture 31/Page 1
Today:
- Torque (10.8, 11.6)
- Newton's second Law of rotation (10.9)
Torque τ (10.8 and 11.6)
Measures the effectiveness of a force causing rotation.


In the general case, when F is not perpendicular to the lever arm r , only the component



of F which is perpendicular to r causes torque. The component parallel to r causes no
torque.
O - the point through which the axis of rotation passes
Red dot – location where the force is applied

r - vector from the origin to point where force is applied. ( r = a.k.a the lever/moment
arm – engineers use instead of “torque” the word “moment”.)


Magnitude: t  rF sin f
(ϕ is the angle between r and F )
So torque is at a max if ϕ = 90º and zero when ϕ = 0º.
  
Torque vector: t  r  F
Direction of torque: use right hand rule
  
t  r1  F1  r1 F1   r1 F1 zˆ
  
t  r2  F2  r2 F2   r2 F2 zˆ
Here, we define out of the page to be positive and into the page negative.
PES 1110 Fall 2013, Spendier
Lecture 31/Page 2
Example 1:
To loosen a pipe, a plumber applies his full weight of 900 N to the end of a cheater by
standing on it. The distance from the center of the fitting to the point where the weight
acts is 0.80 m, and the wrench handle and cheater make an angle of 19º with the
horizontal. Find the magnitude and direction of the torque he applies about the center of
the pipe fitting.
Example 2:
Calculate the net torque about point O for the two forces applied as in the Fig below. The
rod and both forces are in the plane of the page. In what direction will the rod rotate?
Which force will result in the larger torque and why?
PES 1110 Fall 2013, Spendier
Lecture 31/Page 3
Newton’s Second Law of Rotation
Now that we have the rotational analogue to force, we can come up with a rotational form
of Newton’s second law.
Consider
a single mass m on a massless rod of length r, secured at the other end. A force

F is applied to m.
Newton’s 2nd law:
Ft  mat
In rotation, torque plays the role of force, hence if we multiply both sides by r
t  rFt  rmat
Using that rotational acceleration is related to tangential linear acceleration
at  ar
We have
t  rm(ar )   mr 2  a  I a
Hence
t  Ia
(this makes sense since moment of inertia plays the role of mass)
If we had more than one force acting on the mass m, we would need to replace τ with the
net torque (sum of all torques) τnet. Also this equation works on continuous bodies not
just the point particle we have used to show it.
Newton’s Second Law of Rotation:


 t  Ia
net
So the second law of rotation says that addition of all torques cause a change in angular
acceleration α. This amount of change is offset by the body’s amount of inertia I.
This equation is only true for spinning motion (pure rotation) with the origin of
your coordinates at the axis of rotation!
PES 1110 Fall 2013, Spendier
Example 3:
A solid cylinder with moment of inertia 25 kg · m2 and
radius 0.5 m has a rope wrapped around it. The rope is pulled
and the cylinder spins about its center with angular
acceleration 1 rev/s2. What is the tension in the rope? Ignore
friction.
Example 4:
Block 1 has mass m1 = 460 g, block 2 has mass m2 =
500 g, and the pulley, which is mounted on a horizontal
axle with negligible friction, has radius R = 5.00 cm.
When released from rest, block 2 falls 75.0 cm in 5.00 s
without the cord slipping on the pulley. (a) What is the
magnitude of the acceleration of the blocks?
(b) What is the tension T2?
(c) What is the tension T1?
(d) What is the magnitude of the pulley’s angular
acceleration?
(e) What is its rotational inertia?
Lecture 31/Page 4
PES 1110 Fall 2013, Spendier
Lecture 31/Page 5