How Do You Calculate Acceleration and Friction for a Rolling Spherical Shell?

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The discussion focuses on calculating the acceleration, friction force, and minimum coefficient of friction for a hollow spherical shell rolling down a slope at an angle of 38 degrees. The initial calculations for acceleration yielded incorrect results, leading to confusion about the sine function's evaluation. The correct approach involves recognizing that the sine of 38 degrees must be calculated in degrees, not radians, which resolves the discrepancies in the acceleration value. The friction force was also miscalculated, but the method using Newton's second law was correctly applied. Ultimately, the key takeaway is the importance of using the correct units when calculating trigonometric functions in physics problems.
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Q1.) A hollow, spherical shell with mass 1.70 rolls without slipping down a slope angled at 38.0.

-Find the acceleration.
-Find the friction force.
-Find the minimum coefficient of friction needed to prevent slipping.

If it rolls wthout slippng, the increase in Kinetic Energy equals the decrease in Potential Energy. The kinetic energy is in two parts, rotation and translation.
M g H sin 38 = (1/2) M V^2 + (1/2) I (V/R)^2
= [(1/2) + (1/3)] M V^2
The M's and R's cancel and
gH sin 38 = (5/6) V^2
V = sqrt (1.2 g H sin 38) = sqrt (2 a H)
where a is the acceleration
a = 0.6 g sin 38

= 1.74 which is wrong



Use that acceleration and Newton's second law t compute the actual friction force, F.
M g sin 38 - F = M a
Mg sin 38 - M*0.6 g sin 38 = F
F = 0.4 M g sin 38

=1.975 which is wrong

To provide this amount of friction, the static coefficient of friction mu,s must equal or exceed a value given by
M g cos 38 * mu,s = 0.4 M g sin 38
mu,s = 0.4 tan 38

=.124 which is also wrong.

Can someone please check over what i did and see if i miscalculated or something.
 
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Black_Hole? said:
Q1.) A hollow, spherical shell with mass 1.70 rolls without slipping down a slope angled at 38.0.

-Find the acceleration.
-Find the friction force.
-Find the minimum coefficient of friction needed to prevent slipping.

If it rolls wthout slippng, the increase in Kinetic Energy equals the decrease in Potential Energy. The kinetic energy is in two parts, rotation and translation.
M g H sin 38 = (1/2) M V^2 + (1/2) I (V/R)^2
= [(1/2) + (1/3)] M V^2
The M's and R's cancel and
gH sin 38 = (5/6) V^2
V = sqrt (1.2 g H sin 38) = sqrt (2 a H)
where a is the acceleration
a = 0.6 g sin 38

= 1.74 which is wrong

6/5*g*x*sin38 = 2*a*x

a = .6*g*sin38 = .6*9.8*.616 = 3.63 m/s2
 
ok how does the sin(38) = .616?

i keep getting .296369

It is the right answer though
 
You're evaluating the sine in units of radians. Switch to degrees (or multiply 38 with pi/180 before evaluating the sine).
 
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