Problem 642
Question
A uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. The frictional force on the disc is \(\\{\mathrm{A}\\}[(\mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{B}\\}[(2 \mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{C}\\} \mathrm{Mg} \sin \theta\) \(\\{\mathrm{D}\\}\) None
Step-by-Step Solution
Verified Answer
The frictional force on the disc rolling without slipping down an inclined plane at an angle θ with the horizontal is given by \(f = \frac{2Mg\sin\theta}{3}\), which corresponds to option B.
1Step 1: Draw a free body diagram
Draw a free body diagram for the disc while it's rolling down the incline showing all the forces acting on it: gravitational force (Mg), frictional force (f), and normal force (N).
2Step 2: Determine the equations for the horizontal and vertical forces
Resolve the forces in the horizontal (x) and vertical (y) directions, and describe the acceleration and torque acting on the disc:
Sum of the forces in the x-direction: \(f - Mg\sin\theta = Ma\)
Sum of the forces in the y-direction: \(N - Mg\cos\theta = 0\)
3Step 3: Determine the rotational equation
Since the disc is rolling without slipping, we need to consider a rotational equation using the equation of torque about the center of the disc:
Torque = I * angular acceleration
\(fR = \frac{1}{2}MR^2α\)
Where I is the moment of inertia of the disc (\(\frac{1}{2}MR^2\)) and α is the angular acceleration.
4Step 4: Determine the No-slip condition equation
A no-slip condition means that when the disc is rolling, the point of contact with the inclined plane has zero velocity. Using the relationship between linear acceleration and angular acceleration:
\( a = Rα \)
5Step 5: Solve the equations for frictional force
Using Steps 2, 3, and 4, we can solve the system of equations to find the frictional force (f):
From Step 4, \(α = \frac{a}{R}\).
Plugging this value into Step 3 equation: \(fR = \frac{1}{2}MR^2 (\frac{a}{R}) \)
Solving for f: \(f = \frac{1}{2}Ma\)
Now, lets substitute \(f = \frac{1}{2}Ma\) into the Step 2 equation of horizontal forces: \(\frac{1}{2}Ma - Mg\sin\theta = Ma\)
Solving for a: \(a = \frac{2}{3}g\sin\theta\)
Finally, substituting the value of a back to frictional force equation: \(f = \frac{1}{2}M(\frac{2}{3}g\sin\theta)\)
6Step 6: Choose the correct answer
Comparing this result with the given options:
\(f = \frac{2Mg\sin\theta}{3}\), which corresponds to option B.
Key Concepts
Inclined PlaneFrictional ForceRotational Dynamics
Inclined Plane
An inclined plane is a flat surface tilted at an angle relative to the horizontal. It's a classic example used to study dynamics due to its simple yet insightful setup. Imagine a ramp; that's an inclined plane. Objects placed on such a surface experience forces due to gravity acting downwards. However, since the plane is tilted, these forces behave differently compared to forces on a flat surface. By resolving the forces, we can determine which make the object slide down or stay put.
For instance, when a disc rolls down an inclined plane, gravity pulls it downwards. This gravitational force can be split into two components: one perpendicular to the plane (normal force) and one parallel to the plane. The parallel component is given by \( Mg \sin \theta \), where \( \theta \) is the angle of inclination and \( Mg \) is the weight of the disc. This is what primarily causes the object to accelerate down the slope.
Inclined planes are perfect for studying concepts like acceleration and how forces such as gravity and friction interact. They also help explain why certain movements like rolling or sliding occur, making physics tangible.
For instance, when a disc rolls down an inclined plane, gravity pulls it downwards. This gravitational force can be split into two components: one perpendicular to the plane (normal force) and one parallel to the plane. The parallel component is given by \( Mg \sin \theta \), where \( \theta \) is the angle of inclination and \( Mg \) is the weight of the disc. This is what primarily causes the object to accelerate down the slope.
Inclined planes are perfect for studying concepts like acceleration and how forces such as gravity and friction interact. They also help explain why certain movements like rolling or sliding occur, making physics tangible.
Frictional Force
Frictional force is the resistance an object encounters when moving over a surface. It's essential in rolling motion, providing the grip necessary to start and sustain movement. In our example of a disc rolling down an inclined plane, friction is what prevents the disc from simply sliding down. Instead, it allows the disc to roll by providing the torque needed for rotation.
In rolling motion, the friction force does not do work as it doesn't cause energy loss; it merely ensures the "no-slip" condition. It balances the motion, allowing the disc to turn without slipping. This frictional force, expressed as \( f \), acts upwards along the plane, opposing the motion caused by gravity. The frictional force for a rolling object can be calculated if other forces are known, using equations derived from Newton’s laws.
In our exercise, solving for the frictional force involves determining the disc's acceleration and applying it to derive the frictional force formula: \( f = \frac{2}{3} Mg \sin \theta \). This demonstrates how friction fits into the broader dynamics of motion, playing a vital role in real-world applications.
In rolling motion, the friction force does not do work as it doesn't cause energy loss; it merely ensures the "no-slip" condition. It balances the motion, allowing the disc to turn without slipping. This frictional force, expressed as \( f \), acts upwards along the plane, opposing the motion caused by gravity. The frictional force for a rolling object can be calculated if other forces are known, using equations derived from Newton’s laws.
In our exercise, solving for the frictional force involves determining the disc's acceleration and applying it to derive the frictional force formula: \( f = \frac{2}{3} Mg \sin \theta \). This demonstrates how friction fits into the broader dynamics of motion, playing a vital role in real-world applications.
Rotational Dynamics
Rotational dynamics deal with the motion of objects that rotate, explaining how forces and torques interact to produce rotational movement. In the case of a disc rolling down an inclined plane, understanding rotational dynamics is critical in predicting its behavior. Unlike pure sliding, rolling involves both translational and rotational motion, requiring a look at angular quantities.
When the disc rolls without slipping, the point of contact with the plane doesn't slide, but rather 'rolls over'. This is due to the no-slip condition, which leads to the relationship between linear and angular acceleration: \( a = R \alpha \), where \( a \) is the linear acceleration, \( R \) is the radius, and \( \alpha \) is the angular acceleration.
To analyze a rolling disc, we consider both the translational motion along the plane and the rotational motion around its center. Torque, a pivotal concept here, is calculated as the product of the frictional force and the radius (\( fR \)). It acts as the driving force for the angular acceleration. Utilizing the moment of inertia for a disc (\( \frac{1}{2}MR^2 \)), we can derive equations that define how the disc moves and rolls down the plane.
Grasping these principles of rotational dynamics clarifies how objects like wheels, gears, and discs behave in systems, widely influencing engineering and everyday physics.
When the disc rolls without slipping, the point of contact with the plane doesn't slide, but rather 'rolls over'. This is due to the no-slip condition, which leads to the relationship between linear and angular acceleration: \( a = R \alpha \), where \( a \) is the linear acceleration, \( R \) is the radius, and \( \alpha \) is the angular acceleration.
To analyze a rolling disc, we consider both the translational motion along the plane and the rotational motion around its center. Torque, a pivotal concept here, is calculated as the product of the frictional force and the radius (\( fR \)). It acts as the driving force for the angular acceleration. Utilizing the moment of inertia for a disc (\( \frac{1}{2}MR^2 \)), we can derive equations that define how the disc moves and rolls down the plane.
Grasping these principles of rotational dynamics clarifies how objects like wheels, gears, and discs behave in systems, widely influencing engineering and everyday physics.
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