Question

A block of mass m and a cylinder of mass 2m are released on arough inclined plane; inclined at an angle 0 = 45° with thehorizontal. The coefficient of friction for all the contact surfacesis 0.5. the accelerations of the block and cylinder given by ng find the value of n.

Answer 1

\frac{\sqrt{2}}{3}

Answer 2

To determine the acceleration of the block and cylinder, we analyze the forces acting on each object and their motion. We assume that the block and cylinder move together with a common acceleration a down the inclined plane.

Given:

• Mass of block = m
• Mass of cylinder = M = 2m
• Angle of inclination θ = 45°
• Coefficient of friction μ = 0.5 for all contact surfaces.
• The cylinder is a solid cylinder, so its moment of inertia I = (1/2)MR^2.

We set up a coordinate system with the x-axis along the incline (downwards positive) and the y-axis perpendicular to the incline (upwards positive).

1. Analysis of the Cylinder (Mass M = 2m):

• Forces:

  • Gravitational force component parallel to incline: Mg sinθ
  • Friction force from the block f_c (acting up the incline, opposing tendency to slip down)
  • Normal force from the block N_c (perpendicular to the incline)
  • Gravitational force component perpendicular to incline: Mg cosθ

• Equations of Motion:

  • Linear motion (along x-axis): M a = Mg sinθ - f_c 2m a = 2mg sin45° - f_c (Equation 1)
  • Rotational motion (about its center of mass): τ = I α The friction force f_c causes rotation. f_c R = I α For rolling without slipping, a = Rα, so α = a/R. Substitute I = (1/2)MR^2 = (1/2)(2m)R^2 = mR^2: f_c R = (mR^2) (a/R) f_c = ma (Equation 2)
  • Linear motion (along y-axis): N_c - Mg cosθ = 0 (No acceleration perpendicular to the incline) N_c = Mg cosθ = 2mg cos45° = 2mg (1/√2) = √2 mg (Equation 3)

• Calculate acceleration a: Substitute f_c from Equation 2 into Equation 1: 2m a = 2mg sin45° - ma 3m a = 2mg sin45° a = (2/3)g sin45° a = (2/3)g (1/√2) = (√2/3)g

• Check condition for rolling without slipping: The required friction force f_c = ma = m((√2/3)g) = (√2/3)mg. The maximum static friction available is f_{c,max} = μ N_c = 0.5 * (√2 mg) = (√2/2)mg. Since (√2/3)mg < (√2/2)mg (because 1/3 < 1/2), the cylinder can indeed roll without slipping on the block.

2. Analysis of the Block (Mass m):

• Forces:

  • Gravitational force component parallel to incline: mg sinθ
  • Friction force from the cylinder f_{bc} (acting down the incline, by Newton's third law, f_{bc} = f_c = (√2/3)mg)
  • Normal force from the cylinder N_{bc} (acting perpendicular to the incline, into the block, by Newton's third law, N_{bc} = N_c = √2 mg)
  • Friction force from the plane f_b (acting up the incline, opposing motion)
  • Normal force from the plane N_b (perpendicular to the incline)
  • Gravitational force component perpendicular to incline: mg cosθ

• Equations of Motion:

  • Linear motion (along y-axis): N_b - mg cosθ - N_{bc} = 0 N_b = mg cos45° + N_{bc} N_b = mg (1/√2) + √2 mg = (1/√2)mg + (2/√2)mg = (3/√2)mg (Equation 4)
  • Linear motion (along x-axis): m a = mg sinθ + f_{bc} - f_b Substitute a = (√2/3)g and f_{bc} = (√2/3)mg: m ((√2/3)g) = mg sin45° + (√2/3)mg - f_b (√2/3)mg = mg (1/√2) + (√2/3)mg - f_b (√2/3)mg = (1/√2)mg + (√2/3)mg - f_b This simplifies to 0 = (1/√2)mg - f_b So, f_b = (1/√2)mg

• Check condition for sliding on the inclined plane: The required friction force f_b = (1/√2)mg. The maximum friction available from the plane is f_{b,max} = μ N_b = 0.5 * (3/√2)mg = (3/2√2)mg. Since (1/√2)mg < (3/2√2)mg (because 1 < 3/2), the block can slide with this acceleration.

Since both conditions are satisfied, the assumption of common acceleration a = (√2/3)g is valid.

The problem states the acceleration is given by ng. Comparing a = (√2/3)g with a = ng, we find n = √2/3.

The final answer is $\boxed{\frac{\sqrt{2}}{3}}$

Explanation of the solution:

1. Assume a common acceleration a for both the block and the cylinder.
2. For the cylinder, apply Newton's second law for linear motion (down the incline) and rotational motion (about its center). Use the rolling without slipping condition (a = Rα) and the moment of inertia for a solid cylinder (I = (1/2)MR^2). This yields the acceleration a = (2/3)g sinθ and the required friction f_c = ma.
3. Verify that the required friction f_c is less than or equal to the maximum static friction μN_c between the cylinder and the block.
4. For the block, apply Newton's second law for linear motion (down the incline) and perpendicular to the incline. Account for the normal and friction forces exerted by the cylinder on the block (Newton's third law).
5. Verify that the required friction f_b from the inclined plane is less than or equal to the maximum static friction μN_b between the block and the plane.
6. If both friction conditions are met, the calculated acceleration a is correct. The value of n is then determined from a = ng.

Answer: The value of n is $\frac{\sqrt{2}}{3}$.