4 4. This problem has 2 parts. A) A 1.5 N force is applied tangentially for a period of 10 seconds to a soliddisk able to rotate of mass 2.00 kg and radius 0.25 m. VWhat is the final radial velocity? What is the finalkinetic energy? B) The same 1.5 N force is applied towards the center of mass of the same 2.00 kgobject which is now able to move laterally. (in a straight line). What will be its final speed? What will beits final kinetic energy?Part aM=2.00kgR = 0.25 mF = 1.5 NPart bM=2.00kgF = 1.5 N

Assuming the system starts from rest. a) From torque balance at center of the sphere - F×R=…

Answered: 4. This problem has 2 parts. A) A 1.5 N…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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4. This problem has 2 parts. A) A 1.5 N force is applied tangentially for a period of 10 seconds to a solid
disk able to rotate of mass 2.00 kg and radius 0.25 m. VWhat is the final radial velocity? What is the final
kinetic energy? B) The same 1.5 N force is applied towards the center of mass of the same 2.00 kg
object which is now able to move laterally. (in a straight line). What will be its final speed? What will be
its final kinetic energy?
Part a
M=2.00kg
R = 0.25 m
F = 1.5 N
Part b
M=2.00kg
F = 1.5 N

Expert Solution

Step 1

Assuming the system starts from rest.

$F r o m t o r q u e b a l a n c e a t c e n t e r o f t h e s p h e r e - F \times R = I α = \frac{1}{2} M R^{2} α 1.5 \times 0.25 = \frac{1}{2} \times 2 \times 0 . 25^{2} α α = 6 r a d / s^{2} w = w_{0} + α t w = 0 + 6 \times 10 = 60 r a d / s T h e r a d i a l v e l o c i t y i s - V_{r} = r w = 0.25 \times 60 V_{r} = 15 \frac{m}{s} K i n e t i c e n e r g y i s - K . E . = \frac{1}{2} M {V_{r}}^{2} = \frac{1}{2} I w^{2} K . E . = \frac{1}{2} \times 2 \times 15 K . E . = 225 J$

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