Physics, asked by madhu318, 1 year ago

A wheel of radius 1m rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with ground is

Answers

Answered by abhi178
57

Situation is shown in figure. Let initially point of contact with ground is P , then after half revolution, point shift to top of the wheel at P' as you can see in figure.

so, displacement = final position - initial position

= PP' [ see figure ]

= √{length of half revolution² + diameter ²}

length of half revolution = πr = 3.14 × 1 = 3.14 m

diameter = 2r = 2 × 1 = 2m

so, displacement = √{3.14² + 2²}

= √{9.8596 + 4}

= √{13.8596} = 3.72 m

hence, The magnitude of the displacement of the point of the wheel initially in contact with ground is 3.72m

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Answered by sumita65
53

Answer:

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