(ex) A ball of mass 2.0 kg is connected by two massless strings, each with length L = 0.25 m, to a vertical rotating rod. The strings are tied to the rod with separation d = 0.40 m, and are taut. The period of rotation is 0.50 seconds. (a) Determine the tangential speed of the ball. (b) Determine the tensions in the upper and lower strings. (c) Suppose the rotation is slowed, so that the lower string just barely goes slack. Then what is the new tangential speed of the ball? LACOS

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Please check my work and answer part 3. This is an example question NOT HOMEWORK
12) (ex) A ball of mass 2.0 kg is connected by two massless strings, each with length L = 0.25 m, to a
vertical rotating rod. The strings are tied to the rod with separation d = 0.40 m, and are taut. The
period of rotation is 0.50 seconds. (a) Determine the tangential speed of the ball. (b) Determine the
tensions in the upper and lower strings. (c) Suppose the rotation is slowed, so that the lower string just
barely goes slack. Then what is the new tangential speed of the ball?
Tu= upper
Tu = my²
2
र
2² + 1² = 1²
2²+ 1² = 25²
2Rcose
Tu +T+Fq=T + Fg = m ²²
X Tu cose + T cose 10
y Tusine +Tsine. -mg"
1- ma to
2
mg
2sine
c) V dan=2πir = 21T (15)
+
Tv
| mv²
r
map
Vtain=2пrf
ƒ==
눈
2 пс
2πT √2²-2
T
b) Tv=mv²
2RCOSE
T₁=mv²
r
r²+ ()² = 2²
r²=6²-2²
r= -√√₁25² +4²
2RCOSE
Tu = 2(1.88²)
a) v / = 1.88m/s)
ma
2 sin
T=35.2 N
2(1.882)
₁=2(1500336,9
0
Cost
L
CO53=2
20.15)cos369
T=2.52 N
L
=V₁
+
25
6056 =18
20 = 36.9
tan
та
2sing
(9.8)
sin36.9
r = 0.15
2(9.8)
Sin 36.9
Transcribed Image Text:12) (ex) A ball of mass 2.0 kg is connected by two massless strings, each with length L = 0.25 m, to a vertical rotating rod. The strings are tied to the rod with separation d = 0.40 m, and are taut. The period of rotation is 0.50 seconds. (a) Determine the tangential speed of the ball. (b) Determine the tensions in the upper and lower strings. (c) Suppose the rotation is slowed, so that the lower string just barely goes slack. Then what is the new tangential speed of the ball? Tu= upper Tu = my² 2 र 2² + 1² = 1² 2²+ 1² = 25² 2Rcose Tu +T+Fq=T + Fg = m ²² X Tu cose + T cose 10 y Tusine +Tsine. -mg" 1- ma to 2 mg 2sine c) V dan=2πir = 21T (15) + Tv | mv² r map Vtain=2пrf ƒ== 눈 2 пс 2πT √2²-2 T b) Tv=mv² 2RCOSE T₁=mv² r r²+ ()² = 2² r²=6²-2² r= -√√₁25² +4² 2RCOSE Tu = 2(1.88²) a) v / = 1.88m/s) ma 2 sin T=35.2 N 2(1.882) ₁=2(1500336,9 0 Cost L CO53=2 20.15)cos369 T=2.52 N L =V₁ + 25 6056 =18 20 = 36.9 tan та 2sing (9.8) sin36.9 r = 0.15 2(9.8) Sin 36.9
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