The Spinning Disk and Sliding Block

Question

Please help me with this exercise!

A uniform solid disk of mass M = 5 kg and radius R = 0.4 m is mounted on a frictionless horizontal axle through its center, allowing it to rotate freely in a vertical plane. Initially, the disk is at rest. A small block of mass m = 2 kg is placed on the edge of the disk at its highest point (directly above the center). The coefficient of static friction between the block and the disk is μs = 0.6, and the coefficient of kinetic friction is μ = 0.4. At time t = 0, the system is released from rest, and gravity (g = 9.8 m/s2) begins to act.

1. Determine whether the block initially slips or sticks to the disk as the disk begins to rotate.

2. If the block slips, calculate the angular acceleration of the disk and the linear acceleration of the block immediately after release.

3. If the block sticks, calculate the angular acceleration of the disk and the time it takes for the block to reach the lowest point of the disk.

4. Assuming the block slips, find the speed of the block relative to the ground when it reaches the point on the disk that is level with the axle (i.e., at a 90° angle from its starting position).

Thanks!

Buckshot Roulette

Answer 1

So from what I can understand from the problem:

Given:

1. Uniform solid disk:

Mass M = 5 kg, Radius R = 0.4m, Moment of Inertia I_disk = 0.5MR²

1. Block on top of disk:

Mass m = 2 kg (initially at top of the disk (edge, directly above the center)

1. Friction:

Static μ_s = 0.6, Kinetic μ_k = 0.4

1. Gravity : g = 9.8 m/s²
2. Released from rest at t = 0

Part 1. Does the block initially slip or stick?

1. Torque due to the block's weight trying to rotate the disk

τ = mgR

1. This torquw produces an angular acceleration α on the disk:

τ = (I_disk​+mR²) α

mgR=(0.5 MR² + mR²) α

Now plug in the values:

2 x 9.8 x 0.4=(21​ x 5 x 0.42 + 2 x 0.42) α

Calculate:

LHS: 2 x 9.8 x 0.4 = 7.84

RHS: 0.5 x 5 x 0.16 = 0.4

2 x 0.15 = 0.32

Total: 0.4 + 0.32 = 0.72

7.84 = 0.72 α ⇒ α = 0.727/.84​ ≈ 10.89 rad/s²

a_t ​= R α = 0.4 x 10.89 ≈ 4.36m/s²

Normal force = N = mgcos⁡θ

At top (θ=0), N = mg = 19.6 N

Max static friction:

f_max = µ_s N = 0.6 x 19.6 = 11.76N

Required friction to keep it from slipping:

f_req = ma_t = 2 x 4.36 = 8.72 N

Since f_req < f_max​, the block does not slip.

So the final answer would be the block initially sticks to the disk.

Part 2. You can Skip this part because the block does not skip.

Part 3. Angular acceleration and time to reach the lowest point

we know

α = 10.89 rad/s²

It rotates 90°, or θ = π/2 rad

Use rotational kinematics:

θ = ​1/2 α t² ⇒ t = √(2θ/α) ​​≈ √0.2886 ​≈ 0.537s

Final answer 3:

α = 10.89 rad/s²

Time to reach the lowest point ≈ 0.54 s

Part 4. Speed of block when at 90° (bottom right)

Use rotational kinematics again:

ω² = 2αθ = 2 x 10.89 x π/2 = 10.89π ≈ 34.22 ⇒ ω = √34.22 ​≈ 5.85rad/s

Now Speed of block (tangential):

v = ωR = 5.85 x 0.4 ≈ 2.34m/s

Final answer: Speed of the block relative to the ground at 90º = 2.34 m/s