A string is wrapped several times around the rim of a small hoop with radius 8.0

Question

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg . The free end of the string is held in place and the hoop is released from rest (the figure (Figure 1) ). After the hoop has descended 60.0 cm , calculate:

a) the angular speed of the rotating hoop and.

b) the speed of its center.

Item 6 Part A A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the string is held in place and the hoop is released from rest (the figure (Figure 1)). After the hoop has descended 60.0 cm calculate the angular speed of the rotating hoop and rad/s Submit My Answers Give Up Part B the speed of its center 1 of 1 u= m/s Submit My Answers Give Up 0.0800 m

Explanation / Answer

Part A) Here,   PE lost (mgh) = rotational KE gained (1/2I2) + linear KE gained (1/2mv2)

=>   mgh = 1/2(mr2)2 + 1/2mv2

=> gh = r22                                              ( As, v = r)

=> = sqrt(gh)/r

           = sqrt(9.8 * 0.6)/0.08

           = 30.31 rad/sec

=>   the angular speed of the rotating hoop   =   30.31 rad/sec

Part B) speed of its centre =   r *

                                            = 0.08 * 30.31

                                               = 2.424 m/sec

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