Question

A solid cylindrical disk whose mass is 1.4 kg and radius is 8.5 cm, rolls across a horizontal table at a speed of 15 cm/...


A solid cylindrical disk whose mass is 1.4 kg and radius is 8.5 cm, rolls across a horizontal table at a speed of 15 cm/s. What is the instantaneous velocity of the top of the rolling cylinder? What is the angular speed of the rolling disk? What is the kinetic energy of the rolling disk? 

 

Answer

Given:

Mass of the disk is 1.4kg.

Radius of the disk is r=8.5cm=8.5×10-2m.

Rolls across a horizontal table at speed vcom=15cms=15×10-2ms.

Calculation:

The instantaneous velocity at the top of the rolling cylindrical disk is given as:

v=2vcom

Advanced Physics homework question answer, step 2, image 1

Here, vcom is velocity of center of mass given as 15×10-2ms. So,

v=215×10-2msv=0.30ms=30cms

Thus, the instantaneous velocity of the top of the rolling cylinder is 30cms.

The angular speed of the rolling disk is given as:

ω=vcomr

Substitute  for 8.5cm for r and 15cms for vcom.

ω=15cms8.5cmω=1.7647rads

Thus, the angular speed of the rolling disk is 1.7647rads.

The kinetic energy for the rolling disk is given as:

KE=12Icomω2+12mr2ω2KE=12Icomω2+12mvcom2                                                     vcom=ωr

Here, Icom is moment of inertia about center of mass given as:

Icom=12mr2Icom=121.4kg8.5×10-2m2Icom=5.0575×10-3kg m2

Now, kinetic energy:

KE=125.0575×10-3kg m21.7647rads2+121.4kg15×10-2ms2KE=7.8749×10-3+15.75×10-3JKE=23.62×10-3J

Thus, the kinetic energy of the rolling disk is 23.62×10-3J

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