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A disk of radius 25.0cm turns about an axis through the center. The pull on the string produces a linear acceleration a(t)=At on the ball Th
Question
A disk of radius 25.0cm turns about an axis through the center. The pull on the string produces a linear acceleration a(t)=At on the ball The disk starts from rest and after 3 seconds, linear a(3)=1.80m/s2. Find A and then write an expression for the angular acceleration α(t).
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Physics
3 years
2021-08-14T09:49:13+00:00
2021-08-14T09:49:13+00:00 1 Answers
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Answers ( )
Answer:
The value for A is A= 0.6
The angular acceleration is![Rendered by QuickLaTeX.com \alpha (t) = 2.4 \ t \ m/s^2](https://documen.tv/wp-content/ql-cache/quicklatex.com-eb080f6badc43588ab5aa66ab333b842_l3.png)
Explanation:
From the question we are told that
The radius of the disk is![Rendered by QuickLaTeX.com r = 25.0 \ cm = 0.25 \ m](https://documen.tv/wp-content/ql-cache/quicklatex.com-fc5817f6330d81e5425f71ab85b8612b_l3.png)
The linear acceleration is![Rendered by QuickLaTeX.com a(t) = At](https://documen.tv/wp-content/ql-cache/quicklatex.com-e83f049d1fb783274b704ae16f89e308_l3.png)
At time![Rendered by QuickLaTeX.com t = 3 \ s](https://documen.tv/wp-content/ql-cache/quicklatex.com-92d5513e86fa42280cfa149508088614_l3.png)
Generally angular acceleration is mathematically represented as
Now at t = 3 seconds
a(3) = A * 3
=> 1.80 = A * 3
=.> A = 0.6
So therefore
a(t) = 0.6 t
Now substituting this into formula for angular acceleration
substituting for r