Answer:
10.3 s
Explanation:
We are given that
r=43.4 cm=[tex]\frac{43.4}{100}=0.434 m[/tex]
1 m=100 cm
Coefficient of static friction=[tex]\mu_s=0.318[/tex]
Angular acceleration=[tex]\alpha=0.259rad/s^2[/tex]
Initial angular velocity =[tex]\omega_1=0[/tex]
Acceleration of gravity=[tex]g=9.8m/s^2[/tex]
Angular acceleration=[tex]\alpha=\frac{\omega}{t}[/tex]
Using the formula
[tex]0.259t=\omega[/tex]
Force acting on block after time t=[tex]F=m\omega^2 r[/tex]
Friction force=[tex]f=\mu mg[/tex]
[tex]f=F[/tex]
[tex]\mu mg=m\omega^2 r[/tex]
[tex]\mu g=\omega^2 r[/tex]
Substitute the values
[tex]0.318\times 9.8=(0.259t)^2 r[/tex]
[tex]t^2=\frac{0.318\times 9.8}{(0.259)^2\times 0.434}[/tex]
[tex]t=\sqrt{\frac{0.318\times 9.8}{(0.259)^2\times 0.434}}[/tex]
[tex]t=10.3 s[/tex]
Hence, the block will start to slip on the turntable after 10.3 s.
The block will start to slip on the turntable after 10.3s.
Thin is defined as the rate of change of angular velocity with time.
Coefficient of static friction = 0.318
Angular acceleration = 0.259 rad/s²
Initial angular velocity = 0
Acceleration of gravity = 9.8m/s²
Angular acceleration = Angular velocity/ time
0.259t = ω
Force = mω²r
Frictional force = μmg
μmg = mω²r
Substitute the values into the equation
0.318 ₓ 9.8 = (0.259t)²r
t² = (0.318 ₓ 9.8) / (0.259)² ₓ 0.434))
t² = 106.09
t = √106.09 = 10.3s
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