Question

A uniform disk of mass 2kg and radius 10cm is rolling without slipping down a slope that is inclined by 30° with the horizontal. a) Find the disk's linear acceleration. b) Find the minimum coefficient of static friction between the disk and ramp for which the disk can roll without slipping?

Answer 1

a) The disk's linear acceleration is approximately2.45 ,
`\text{m/s}^2\)`. b) The minimum coefficient of static friction between the disk and the ramp for which the disk can roll without slipping is0.20.

Step-by-step explanation:

a) The linear acceleration a of the rolling disk can be determined using the rotational dynamics equation
`\(a = (r \alpha)/(1 + (r^2)/(k^2))\)`, where r is the radius and
`\(\alpha\)` is the angular acceleration. In the case of rolling without slipping,
`\(\alpha = (a)/(r)\)`. The linear acceleration is given by
`\(a = g \sin(\theta)\)`, where g is the acceleration due to gravity and
`\(\theta\)` is the angle of the slope. Substituting these values, we find
`\(a \approx 2.45 \, \text{m/s}^2\)`.

b) The minimum coefficient of static friction
`(\(\mu_s\))` can be found using the condition for rolling without slipping, which is
`\(f_s \leq \mu_s N\)`, where
`\(f_s\)` is the static friction force and N is the normal force. The normal force can be calculated as
`\(N = mg \cos(\theta)\)`, and the static friction force is
`\(f_s = \mu_s N\)`. At the threshold of slipping, the static friction force equals the component of gravitational force parallel to the slope, i.e.,
`\(f_s = mg \sin(\theta)\)`. Substituting these values, we get
`\(\mu_s \geq \tan(\theta)\), and for \(\theta = 30^\circ\), \(\mu_s \geq 0.20\)`.

In summary, the linear acceleration of the rolling disk is approximately
`\(2.45 \, \text{m/s}^2\)`, and the minimum coefficient of static friction required for rolling without slipping is
`\(0.20\)`.