Tutorial Exercise Find the net torque on the wheel in the figure below about the axle through O perpendicular to the page, taking a = 16.0 cm and b = 32.0 cm. 12.0 N 30.0% = 10.0 N 9.00 N Step 1 The total torque on an object is the sum of the individual torques. Remember that d is the perpendicular distance from the rotation axis to the line of action of the force F, called the moment arm or lever arm of F. We have the following. ΣT=Fd N)(0.32 m) N)(0.16 m)-(10.0 N)(1 N m e m This represents a torque vector that points into the plane of the page, and results in a counterclockwise turning by the force.
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.
![### Tutorial Exercise
**Objective:** Find the net torque on the wheel in the figure below about the axle through \( O \) perpendicular to the page, taking \( a = 16.0 \, \text{cm} \) and \( b = 32.0 \, \text{cm} \).
**Diagram Description:**
- A wheel is shown with forces acting on it.
- Three forces are applied:
- \( 10.0 \, \text{N} \) horizontally to the right at the top.
- \( 12.0 \, \text{N} \) at a 30.0° angle below the horizontal on the left.
- \( 9.00 \, \text{N} \) vertically downward on the right.
- The distances \( a \) and \( b \) represent radii from point \( O \) to where the forces are applied.
### Step 1
**Explanation:**
The total torque on an object is the sum of the individual torques. Remember that \( d \) is the perpendicular distance from the rotation axis to the line of action of the force \( \vec{F} \), called the moment arm or lever arm of \( \vec{F} \). We calculate torque using the formula:
\[
\sum \tau = \sum Fd
\]
**Equation for Total Torque:**
\[
= + \left( \underline{\quad} \, \text{N} \right)(0.16 \, \text{m}) - (10.0 \, \text{N})(\underline{\quad} \, \text{m}) - (\underline{\quad} \, \text{N})(0.32 \, \text{m})
\]
\[
= \underline{\quad} \, \text{N} \cdot \text{m}
\]
This represents a torque vector that points into the plane of the page, resulting in a **counterclockwise** turning by the force.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F389450d8-f111-499a-811c-3e6b049ffcf7%2Fbaf7fe70-99d5-42a3-931c-4f9c0bfa9c91%2Ffu1kk2s_processed.jpeg&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
