# 基礎物理題目 請幫幫忙

1. The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 12.0 sec. At this point, someone opened the lid and the safety switch turns off the washer. The tub slows to rest in 14.0s. Through how many revolution does ths tub turn during this 26s interval? Assume contact angular acceleration while it is starting and stopping.

2. What is the tangential acceleration of a bug on the rim of a 10.0 inch diameter dist if the dist moves from rest to an angular speed of 75 rev/min in 5 sec? When the dist is at its final speed, what is the tangential velocity of the bug? one second after the bug starts from the rest, what is its tangential acceleration? centripetal acceleration? total acceleration and angle?

Thanks

Rating

1.

ω=ω0+αt

5=0+α*12

α=5/12

θ=ω0t+(1/2) αt2

=0+(1/2)*( 5/12)*122

=30(rev)

ω=ω0-αt

0=5-α*14

α=5/14

θ=ω0t-(1/2) αt2

=5*14-(1/2)*( 5/14)*142

=35(rev)

So,the tub turns 30+35=65(rev) during the 26s interval.

2.

ω=75 rev/min=1.25 rev/s

ω=ω0+αt

1.25=0+α*5

α=0.25

at=rα

=10*0.25

=2.5(inch/s2)

v=rω

=10*1.25

=12.5(inch/s)

ω=ω0-αt

0=1.25-α*1

α=1.25

at=rα

=10*1.25

=12.5(inch/s2)

ac=rω2

=10*1.252

=15.625(inch/s2)

a=√(at2+ac2)

=√(12.52+15.6252)

=20.0(inch/s2)

Source(s): 以上為個人淺薄意見，若有錯誤，歡迎糾正
• Anonymous
7 years ago

我每次都是去這里看的哦, http://Lvmiss。com

勘仱咇互