Question

A drum of 4-in radius is attached to a disk of 8-in radius. The disk and drum have a total weight of 10 lb and a combined radius of gyration of 6 in. A cord is attached as shown and pulled with a force P of magnitude 5 lb. Knowing that the disk rolls without sliding, determine:

(a) The angular acceleration of the disk and the acceleration of gravity center.
(b) The minimum value of the coefficient of static friction compatible with this motion

A:
A) Angular acceleration of the disk = P/I​, Acceleration of gravity (g) = 2R/3
B) Angular acceleration of the disk = 2P/I, Acceleration of gravity (g) = 4R/3​
C) Angular acceleration of the disk = 3P/I​, Acceleration of gravity (g) = 6R/3
D) Angular acceleration of the disk = P/2I​, Acceleration of gravity (g) = R/3

B:
A) μs≥1/2
B) μs≥2/3
C) μs≥3/4
D) μs≥1/3

Answer 1

A drum of 4-in radius is attached to a disk of 8-in radius. The disk and drum have a total weight of 10 lb and a combined radius of gyration of 6 in. A cord is attached as shown and pulled with a force P of magnitude 5 lb. Knowing that the disk rolls without sliding,

A: Angular acceleration of the disk =
`\( (P)/(I) \)`, Acceleration of gravity
`(\( g \)) = \( (2R)/(3) \)`

B:
`\( \mu_s \geq (2)/(3) \)`

Step-by-step explanation:

**(A) Angular Acceleration and Acceleration of Gravity Center:**

To find the angular acceleration
`(\(\alpha\))` , we use the formula
`\(\tau = I\alpha\)`, where
`\(\tau\)` is the torque,
`\(I\)` is the moment of inertia, and
`\(\alpha\)` is the angular acceleration. The torque is given by
`\(\tau = r * F\),` where
`\(r\)` is the radius and
`\(F\)` is the force applied. For the disk rolling without sliding,
`\(I = (MR^2)/(2)\),` where
`\(M\)` is the mass of the disk. The total moment of inertia for the system is the sum of the moments of inertia of the drum and disk.

The angular acceleration
`(\(\alpha\))` is given by
`\( \alpha = (\tau)/(I) \)` , and the acceleration of the gravity center is
`\(g = (2R)/(3)\).`

**(B) Minimum Coefficient of Static Friction:**

For rolling without sliding, the condition for static friction
`(\(\mu_s\))` is
`\( \mu_s \geq (r)/(R) \),` where
`\(r\)` is the radius of the drum in contact with the surface. In this case,
`\(r = (R)/(2)\), so \( \mu_s \geq (2)/(3) \).`

In summary, option A is the correct expression for angular acceleration and gravity center acceleration, and option B correctly represents the minimum coefficient of static friction compatible with the motion.