A thin, light string is wrapped around the rim of a 3.90 kgsolid uniform disk (s

Question

A thin, light string is wrapped around the rim of a 3.90 kgsolid uniform disk (similar to a puck on ice without friction) that is 31.0 cm in diameter. A person pulls on the string with a constant force of 101.5 N tangent to the disk, as shown in the figure below (Figure 1) . The disk is not attached to anything and is free to move and tur

A) Find the angular acceleration of the disk about its center of mass.

B) Find the linear acceleration of its center of mass

C) If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the angular acceleration of the disk about its center of mass?

D) If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the linear acceleration of its center of mass?

Explanation / Answer

1) angular acceleration = torque/moment of inertia.

Torque = rF sin90 = rF ( sin90 = 1 )

= 0.155 × 101.5

= 15.7325

Moment of inertia = 1/2MR^2

= 0.5*3.90*0.155^2

= 0.04684

:- angular acceleration = 15.7325/0.04684

= 335.877 rad/s^2.

2) liners acceleration = 0.155 × 335.877

= 52.06 m/s^2.

3) when disk is replace by hollow cylinder moment of inertia becomes double.

Moment of inertia = 335.877/2

= 167.9385rad/s^2

4) linear acceleration = F/M

= 101.5/3.90

= 26.02m/s^2.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.