A uniform disk with mass 41.8kg and radius 0.280m is pivoted

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A uniform disk with mass 41.8kg and radius 0.280m is pivoted at its center about a horizontal,frictionless axle that is stationary.The disk is initially at rest,and then a constant force 28.0N is applied tangent to the rim of the disk.
1.What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.200 revolution?
2.What is the magnitude α of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.200 revolution?
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一个均衡圆盘，41.8千克，半径0.280米，中心有水平、无摩擦轴力的转动轴。圆盘最初是处于静止状态，然后盘缘施加一个恒定的相切的力，大小28.0N。
1）在0.200圈后，相切方向上的速度大小为多少？
2）在0.200圈后，总加速度的大小为多少？

看不懂,翻一下,要准确.
再问：一个均衡圆盘，41.8千克，半径0.280米，中心有水平、无摩擦轴力的转动轴。圆盘最初是处于静止状态，然后盘缘施加一个恒定的相切的力，大小28.0N。 1）在0.200圈后，相切方向上的速度大小为多少？ 2）在0.200圈后，总加速度的大小为多少？
再答： (1)圆盘的转动惯量：I=mr²/2 圆盘的角加速度：β=Fr/I=2F/mr 根据：ω²－ω0²=2βθ得，0.2圈后的角速度：ω=√2βθ=2√(0.4πF/mr) 所以，此时切向速度：v=ωr=2√(0.4πFr/m) (2)此时的切向加速度：a1=βr=2F/m，法向加速度：a2=ω²r=0.16πF/m 总加速度大小：a=(a1²+a2²)^½ 具体数值你代入数据算一下。不明白再追问
再问： a2=ω²r=0.16πF/m的ω²r和0.16πF/m不等诶……
再答：怎么不相等啊。ω²=0.16πF/mr ω²r=0.16πF/m 么这里求得是圆盘边沿上某点的加速度。如果所求的点不在边缘上，把那个半径换成这一点距离轴的距离就行了。

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