Suppose the student had jumped on to the rim of the merry-go-round without transferring any angular momentum to the merry-go-round. (a) What was the angular speed of the merry-go-round before the student jumped on? Recall that with the child standing at the outer edge, the angular speed is 2.1 rad/s. rad/s (b) By how much did the kinetic energy of the system change when the student jumped on? ΔΚ :
Angular Momentum
The momentum of an object is given by multiplying its mass and velocity. Momentum is a property of any object that moves with mass. The only difference between angular momentum and linear momentum is that angular momentum deals with moving or spinning objects. A moving particle's linear momentum can be thought of as a measure of its linear motion. The force is proportional to the rate of change of linear momentum. Angular momentum is always directly proportional to mass. In rotational motion, the concept of angular momentum is often used. Since it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics. To understand the concept of angular momentum first we need to understand a rigid body and its movement, a position vector that is used to specify the position of particles in space. A rigid body possesses motion it may be linear or rotational. Rotational motion plays important role in angular momentum.
Moment of a Force
The idea of moments is an important concept in physics. It arises from the fact that distance often plays an important part in the interaction of, or in determining the impact of forces on bodies. Moments are often described by their order [first, second, or higher order] based on the power to which the distance has to be raised to understand the phenomenon. Of particular note are the second-order moment of mass (Moment of Inertia) and moments of force.
![A force of
= (3.00î + 6.00j) N is applied to an object that is pivoted about a fixed axis aligned along
the z coordinate axis. The force is applied at a point located at r = (2.00î + 3.00j) m. Find the torque 7
applied to the object.
SOLVE IT
Conceptualize Given the unit-vector notations, think about the directions of the force and position
vectors. If this force were applied at this position, in what direction would an object pivoted at the origin
turn?
Categorize Because we use the definition of the cross product discussed in this section, we categorize
this example as a substitution problem.
Set up the torque vector using the
following equation:
7 = rxF = [(2.00î + 3.00j) m] x [(3.00î + 6.00j) N]
Perform the multiplication:
7 = [(2.00)(3.00)î × î + (2.00)(6.00)î × j
+ (3.00)(3.00)j x î + (3.00)(6.00)j xj] N• m
Evaluate the various terms:
7 = [0 + 12k + -9k + 0] N .m =
kN: m
Notice that both r and F are in the xy plane. As expected, the torque vector is perpendicular to this
plane, having only a z component.
MASTER IT
HINTS:
GETTING STARTED | I'M STUCK!
Suppose the same magnitude force is applied at the same point as in the example, and the torque is
found to have the same magnitude but in the opposite direction of the torque found there. What are the
components of the force?
5) N](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F22499db0-d22f-4756-b954-120799dc3fe6%2Fd6064945-1fae-4946-b53a-5a87877b901a%2Fy84jmu9_processed.jpeg&w=3840&q=75)
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