A thin, light wire is wrapped around the rim of a wheel, as shown in the followi

Question

A thin, light wire is wrapped around the rim of a wheel, as shown in the following figure. The wheel rotates without friction about a stationary horizontal axis that passes through the center of the wheel. The wheel is a uniform disk with radius 0.292 m . An object of mass 4.35 kg is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration

Part A

If the suspended object moves downward a distance of 2.70 m in 2.08 s , what is the mass of the wheel?

Explanation / Answer

2.70 = 0.5 a( 2.08)^2

a ( accelertion),= 1.248 m/s^2 apprx

v( fiinal velocity) = 1.248 (2.08) = 2.59 m/s apprx

mgh = 1/2mv^2 + 1/4MR^2(v/R)^2 ---using conseration of energy ( where mR^2/ 4 is the momnet of i ertia of disk about diamtere)

4.35(9.8) (2.7) = 0.5 ( 4.35) ( 6.74) + 1/4 M( 0.292)^2 ( 2.59/0.292)^2

115.101= 14.6595 + 1.677 M

M = 60.253 kg

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