佳礼资讯网

 找回密码
 注册

ADVERTISEMENT

搜索
楼主: 辉文

物理讨论专区

   关闭 [复制链接]
发表于 2-8-2010 12:57 AM | 显示全部楼层
本帖最后由 vick5821 于 2-8-2010 04:24 PM 编辑

[img][/img]
help pls..thanks
回复

使用道具 举报


ADVERTISEMENT

发表于 2-8-2010 07:01 AM | 显示全部楼层
不知为什么,看不到你上传的图片。
回复

使用道具 举报

发表于 2-8-2010 04:19 PM | 显示全部楼层
nvm..that one i will upload soon..help me with this..

A solid sphere of radius 15.0 cm and mass 10.0kg rolls down an inclined plane make an angle of 25 degree to the horizontal. If the sphere rolls without slipping from rest to the distance of 75.0 cm and the inclined surface is smooth, calculate
a) the total kinetic energy of the sphere
b) the linear speed of the sphere
c)the angular speed about the centre of mass.
(Given the moment of inertia of solid sphere is Icm = 2/5mR^2 and g = 9,81)

I get the answer as
a) 43.76J
b) 2.5 ms-1
c) 16.67

but the teacher said is wrong wor..can help?? thanks
回复

使用道具 举报

发表于 2-8-2010 05:50 PM | 显示全部楼层
回复 383# vick5821


Okay, I just wrote up my solution without calculation. This is your job to complete those calculation.
a) The gain of "total" kinetic energy is equal to the loss of potential energy, that is

Total kinetic energy = mgh, where h = ???

b) Total kinetic energy = 0.5 mv^2 + 0.5I(omega)^2 = 0.5mv^2 + 0.5(0.4mR^2)(v/R)^2, where omega = v/R. Thus, v = ???

c) omega = v/R = ???
回复

使用道具 举报

发表于 2-8-2010 05:52 PM | 显示全部楼层
Another thing to remind you: You have to put the unit for your angular speed.
回复

使用道具 举报

发表于 2-8-2010 07:05 PM | 显示全部楼层
but for a, the total kinetic energy isnt that need to add rotational KE and translation KE??
回复

使用道具 举报

Follow Us
发表于 2-8-2010 07:41 PM | 显示全部楼层
Recall the principle of conservation of energy, which you had learned since your PMR.
回复

使用道具 举报

发表于 2-8-2010 08:19 PM | 显示全部楼层
but then I understand that..but total KE for the sphere suppose to have translation and rotation??
回复

使用道具 举报


ADVERTISEMENT

发表于 2-8-2010 08:33 PM | 显示全部楼层
回复 388# vick5821


   Yes.
回复

使用道具 举报

发表于 2-8-2010 08:37 PM | 显示全部楼层
but then I understand that..but total KE for the sphere suppose to have translation and rotation??
回复

使用道具 举报

发表于 2-8-2010 08:37 PM | 显示全部楼层
Then suppose should get what I get??? if i use conservation of energy then get diff ans wor
回复

使用道具 举报

发表于 2-8-2010 08:45 PM | 显示全部楼层
本帖最后由 vick5821 于 2-8-2010 08:48 PM 编辑

but then I understand that..but total KE for the sphere suppose to have translation and rotation??why use conservation of energy..if use conservation energy then the KE found is the combination of both??
回复

使用道具 举报

发表于 2-8-2010 08:57 PM | 显示全部楼层
You have to get through it yourself; I have gave my explanation clearly.
回复

使用道具 举报

发表于 2-8-2010 09:03 PM | 显示全部楼层
when the body rools down got linear acceleration and angular acceleration rite??
回复

使用道具 举报

发表于 2-8-2010 09:06 PM | 显示全部楼层
“the sphere rolls without slipping from rest
回复

使用道具 举报

发表于 2-8-2010 09:14 PM | 显示全部楼层
this is my working..you see can I do it like this or not..see whether I understand correct or not
a) K(total) = K(translation) + K ( rotational)
                =1/2mv^2 + 1/2 Iw^2              v calculated ffrom the linear motion equation and get 2.5 ms-1 so w= v/r =16.67

then i sub into the formula and get 43.76 J .Can I do it like this???
回复

使用道具 举报


ADVERTISEMENT

发表于 2-8-2010 09:21 PM | 显示全部楼层
Your v cannot be computed from linear motion equation, I can guarantee that. Your potential energy does not be converted entirely to kinetic energy.
回复

使用道具 举报

发表于 2-8-2010 09:25 PM | 显示全部楼层
you mean in my working the PE is not all converted to KE?and why v cannot be clculated from linear equation?
回复

使用道具 举报

发表于 2-8-2010 09:29 PM | 显示全部楼层
Sorry, what I mean "kinetic energy" refer to "translational" kinetic energy.
回复

使用道具 举报

发表于 2-8-2010 09:32 PM | 显示全部楼层
If you want to apply linear motion equation, you have to make sure that all the potential energy is converted entirely to "translational" kinetic energy. (I thought you know people would implicitly mean "kinetic energy" as "translational kinetic energy", "total kinetic energy" as "rotational + translational".)
回复

使用道具 举报

您需要登录后才可以回帖 登录 | 注册

本版积分规则

 

所属分类: 欢乐校园


ADVERTISEMENT



ADVERTISEMENT



ADVERTISEMENT

ADVERTISEMENT


版权所有 © 1996-2026 Cari Internet Sdn Bhd (483575-W)|IPSERVERONE 提供云主机|广告刊登|关于我们|私隐权|免控|投诉|联络|脸书|佳礼资讯网

GMT+8, 31-3-2026 07:39 PM , Processed in 0.086670 second(s), 19 queries , Gzip On.

Powered by Discuz! X3.4

Copyright © 2001-2021, Tencent Cloud.

快速回复 返回顶部 返回列表