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 m
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/16 07:38:50
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 m
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?
需翻译请回复
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?
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 m
看不懂,翻一下,要准确.
额 这个题呀,我试着翻译一下。 一个重41.8kg半径0.280米的圆盘水平放置在平面上,绕中轴旋转,无摩擦,盘子最初是静止的,然后再盘子边缘施加一个28N的切向力。
问:1,当盘子转过0.2转后,边缘上的点切向加速后的速度多大。2.当盘子转过0.2转后,其边缘上点的合成加速度多大。。 有点乱 “of”把我乱的。。。。
额 上面仁兄已经做出来了 不过高中木有学过转...
全部展开
额 这个题呀,我试着翻译一下。 一个重41.8kg半径0.280米的圆盘水平放置在平面上,绕中轴旋转,无摩擦,盘子最初是静止的,然后再盘子边缘施加一个28N的切向力。
问:1,当盘子转过0.2转后,边缘上的点切向加速后的速度多大。2.当盘子转过0.2转后,其边缘上点的合成加速度多大。。 有点乱 “of”把我乱的。。。。
额 上面仁兄已经做出来了 不过高中木有学过转动惯量吧。。
收起
需要用的公式有:M = J α,M = F r,J = m r^2/2, v = ω r, ω^2 -ω0^2 = 2α(θ - θ0),
at = rα, an = r ω^2, a^2 = at^2+an^2,
现说明之:M是力矩,J是转动惯量, α是角加速度,F是切向力为已知,r是半径,ω是角速度,ω0是初始角速度,为0,θ - θ0是角位移等于0.200计算时要化...
全部展开
需要用的公式有:M = J α,M = F r,J = m r^2/2, v = ω r, ω^2 -ω0^2 = 2α(θ - θ0),
at = rα, an = r ω^2, a^2 = at^2+an^2,
现说明之:M是力矩,J是转动惯量, α是角加速度,F是切向力为已知,r是半径,ω是角速度,ω0是初始角速度,为0,θ - θ0是角位移等于0.200计算时要化成弧度,θ0为0,at是切向加速度,an是法向加速度,a是总加速度。
先求出 α,角加速度,再算出ω,角速度,则v可算出,最后去计算加速度。
收起