A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25° with the horizontal.What is the forward speed of the disk after it has rolled 3.0 m,measured along the plane?
A)2.0 m/s
B)3.5 m/s
C)4.1 m/s
D)5.7 m/s
E)6.3 m/s

Question

Quizplus · Accepted Answer

We can use conservation of energy to solve this problem. The final kinetic energy of the disk is equal to the initial potential energy, so we can write:

1/2 mv^2 = mgh

where m is the mass of the disk, v is its final speed, g is the acceleration due to gravity, and h is the height the disk has descended. We need to determine h in terms of the distance the disk has traveled, which we'll call d.

First, we can find the distance traveled parallel to the plane using the angle of the plane and the distance traveled perpendicular to the plane:

sin(25°) = h/d
h = d sin(25°)

Next, we can find the height lost by the disk due to its motion by considering the distance it has rolled assuming it is a solid disk. The circumference of a circle is 2πr, so the distance traveled is:

d_roll = (2πr)/4 = πr/2

where we used the fact that the disk rolls without slipping, so its linear speed at the edge is equal to its rotational speed. We can also write the distance traveled in terms of the height lost:

d_roll = h_loss / sin(25°)

Setting these two expressions for d equal to each other and solving for h_loss, we get:

h_loss = (πr/2) sin(25°)

Now we can substitute this expression for h_loss into our energy equation:

1/2 mv^2 = mg( d sin(25°) - πr/2 sin(25°) )

Solving for v, we get:

v = sqrt(2g( d sin(25°) - πr/2 sin(25°) ))

Substituting in the given values for g, d, and r, we get:

v ≈ 4.1 m/s

So the answer is (C).

A Uniform Solid Disk Is Released from Rest and Rolls